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Full text of "Chemical problems and reactions : to accompany Stöckhardt's Elements of chemistry"



LIBRARY 



UNIVERSITY OF NORTH CAROLINA 



Alcove Shelf 



UNIVERSITY OF N.C. AT CHAPEL HILL 



00011546124 



CHEMICAL AnJ^^y 
PEOBLEMS AND REACTIONS, 

TO ACCOMPANT 

STOCKHARDT'S ELEMENTS OF CPIEMISTRY. 



BY 

JOSIAH P.'^OOKE, Jr. 



#H:?-L:i3fary 

il;^ Vnhimliy of North Ca o;;;.. 



CAMBEIDGE: 

PUBLISHED BY JOHN BAETLETT, 

ISoofeseller to tje Sinfbcrsitj). 

1857. 



C A 31 B E I D G E : 

ELECTROTTPED AND PRINTED BY METCALF AND COMPANY. 



PREFACE. 



This book has been prepared solely for the use of the 
undergraduates of Harvard College. It contains a collection 
of chemical problems and reactions, with references to the sec- 
tions of Stockhardt's Elements of Chemistry, and also a few 
chapters on the chemical nomenclature and the use of chemical 
symbols, subjects which are not sufficiently developed in that 
text-book for the purposes of college instruction. In writing 
chemical symbols the author has adopted a uniform system 
throughout the volume, which, as he hopes, will be found to 
be at once expressive and clear. The problems and reactions 
cover the Inorganic portion of Stockhardt's Elements ; the 
problems have only been extended to the section on the Heavy 
Metals. Beyond this, the reactions alone have been given, as it 
was supposed that, before reaching this section, the student 
will easily be able to propose problems for himself. In 
solving many of the problems it will be found convenient to 
use logarithmic tables of four places, which, with several other 
tables, will be found at the end of the volume. The student 
is advised to remove the tables of logarithms, and paste them 
for use on a card. The difficulty of insuring complete accu- 
racy in the printing of chemical formulge can be known only 
to those who have had to see a book of this kind through the 
press. Several errors have been already discovered, and 
corrected, but others unquestionably exist. 

Cambeidge, May I5th, 1857. 



Digitized by the Internet Archive 

in 2012 with funding from 

University of North Carolina at Chapel Hill 



http://archive.org/details/chemicalproblemsOOcook 



NOMENCLATURE OF CHEMISTRY. 



Origin of the Nomenclature. — Previous to the year 1787 
the names given by chemists or alchemists to substances 
were not conformed to any general rules. Many of these old 
names, such as Oil of Vitriol, Calomel, Corrosive Sublimate, 
Red-Precipitate, Saltpetre, Liver of Sulphur, Cream of Tartar, 
Glauber^s and Epsom Salts, are stUl retained in common use. 
As chemical science advanced, and the number of known 
substances iucreased, it became important to adopt a scientific 
nomenclature. The admirable system now in use is due al- 
most entirely to Lavoisier, who reported to the French Acad- 
emy on the subject, in the name of a committee, in 1787. 
This system, now known as the Lavoisierian nomenclature, 
was generally adopted by scientific men soon after its pub- 
lication, and has not been materially modified since. In it 
the name of a substance is made to indicate the composi- 
tion. 

Names of the Elements. — The names of the elements are 
the only ones which are now independent of any rule. Those 
which were known before the adoption of the nomenclature, 
1 



2 NOMENCLATURE OF CHEMISTRY. 

such, as Sulphur, Phosphorus, Iron, Lead, retain their old 
names. Several of the more recently discovered elements 
have been named in allusion to some prominent property or 
some circumstance connected with their history; as, Oxygen 
from o^vs, yei/mc* (acid-generator) ; Hydrogen, from vbap, yewaca 
(water-generator) ; Chlorine, from -xkapos (green) ; Iodine, 
from labrjs (violet) ; Bromine, from ^papos (fetid odor), &c. 
The names of the newly discovered metals have a common 
termination, um, as Platinum, Potassium, Sodium ; and the 
names of a class of the metalloids terminate in ine, as Chlo- 
rine, Bromine, &c. ; but except in these respects the names 
of the elements are entirely arbitrary. 

Classification of Compounds. — There are three orders of 
chemical compounds : — 1st, Binary Compounds, consisting of 
two elements, or of the representatives of two elements ; 2d, 
Ternary Compounds, consisting of three elements, or of their 
representative ; and Sd, Quaternary Compounds, consisting of 
four elements, or their representatives. There are some chem- 
ical compounds containing more than four elements ; but in 
most cases two or more of these elements are representatives, 
i. e. occupy the place, of only one, as will be explained farther 
on. Binary compounds are subdivided into two classes, Elec- 
tro-Positive Binaries, or Bases, and Electro-Negative Binaries, 
or Acids. Each of these classes is distinguished by a peculiar 
set of properties, or at least this is the case with the promi- 
nent members of either class ; but the two classes merge so 
gradually into each other, that it is impossible to draw a line 
of demarcation between them ; and there is a large class of 
intermediate compounds, which either partake of the proper- 



NOMENCLATURE OF CHEMISTRY. 3 

ties of both, or are entirely indifferent. Indeed, the binary 
compounds may best be regarded as forming a continuous series 
of substances, varying in their properties from those of strong 
acids on the one hand to those of strong bases on the otlier, 
and with every possible grade of qualities between the two 
extremes. In this series each binary may be considered as 
an electro-positive compound or base towards all those which 
precede it, and as an electro-negative compound or acid to- 
wards all those which follow it. Ternary compounds are gen- 
erally, at least in Inorganic Chemistry, composed of two 
binaries, i. e. of an acid and a base, and are then called 
Salts. The quaternary compounds are generally composed of 
two salts, and are called Double Salts. 

Names of Binaries. — The most important binaries, as well 
as those which have been the best studied, are the compounds 
of oxygen with the other elements. To these the generic 
term Oxide has been applied. The electro-positive binaries 
are called simply Oxides of the elements of which they con- 
sist. Thus we have 

Oxide of Hydrogen, consisting of oxygen and hydrogen. 
Oxide of Potassium, " " " potassium. 

Oxide of Sodium, " " sodium. 

When the name of the metal ends in iim, the name of the 
compound with oxygen is frequently formed by changing this 
termination into a, with such other modifications of the termi- 
nal letters as euphony may require. Thus we use, instead of 

Oxide of Sodium, Soda. 

Oxide of Potassium, Potassa. 



4 NOMENCLATURE OF CHEMISTRT. 

Oxide of Calcium, Calcia (or Lime).* 

Oxide of Barium, Baryta. 

Oxide of Strontium, Strontia. 

Oxide of Magnesium, Magnesia. 

Oxide of Aluminum, Alumina. 

The two names are in all cases synonymous. Generally 
oxygen combines with an element in more than one propor- 
tion ; then, in order to distinguish between the different oxides 
of the same element, we use various Latin and Greek prefixes, 
such as suh, proto, sesqui, deuto, hyper. This is well illustrated 
by the names of the different oxides of mercury and man- 
ganese, which are as follows. 

Composition. 

Names. 

Suho:side of Mercury 
Protoxide of Mercury 

Protoxide of Manganese 
Sesquioxide of Manganese 
£li/pero:side of Manganese 

The electro-negative binary compounds of oxygen (the acids) 
are named on a different principle. These are called different 
kmds of acids. If the element forms but one acid with oxy- 
gen, the name is formed by adding to the name of the acid the 
termination ic, with such changes of the final letters as eu- 
phony may require. Thus, carbon and oxygen form Carbonic 

* The common name Lime is much more frequently used than either of its 
scientific synonymes, Oxide of Calcium, or Calcia. Indeed, the last has never 
been in general use. 



Mercury. 


Oxygen. 


100 


4 


100 


8 


Manganese. 


Oxygen. 


27.6 


8 


27.6 


12 


27.6 


16 



Arsenic. 


Oxygen. 


75 


24 


75 


40 


Phosphorus. 


Oxygen. 


32 


24 


32 


40 



NOMENCLATURE OF CHEMISTRY. 

Acid. When the element forms two acids, by combining with 
different amounts of oxygen, the termination ie is reserved for 
that containing the most oxygen, while the termination ous is 
given to the other. We have, for example, 

ArsenioMs Acid 
Arsem'c Acid 

Phosphorotw Acid 
Phosphorzc Acid 

. If oxygen combines with an element in more than two pro- 
portions, to form acids, the names are formed with the Greek 
prefix Jiypo, indicating a less, or the Latin prefix per, indicating 
a greater, amount of oxygen than that contained in the acids 
to whose names they are prefixed. The acid compounds of 
sulphur and oxygen are 

Sulphur. Oxygen. 

IIyposVi\]Amvous Acid 16 8 

Sulphuroits Acid 16 16 

Hypo&\}\^\mxic Acid 16 20 

Sulphur^c Acid 16 24 

The acid compounds of chlorine and oxygen are 

Chlorine. Oxj'gen. 

IIypoQh\.OYous Acid 35.5 8 

Chlovous Acid 
Jlypochloric Acid 
Chloric Acid 
PerdiAoTic Acid 

Very frequently the higher degrees of oxidation of an ele- 
1* 



35.5 


24 


35.5 


.32 


35.5 


40 


35.5 


56 



Nitrogen. 

14 


Oxygi 
8 


14 


16 


14 


24 


14 


32 


14 


40 



b NOMENCLATURE OF CHEMISTRY. 

ment are acids, when the lower degrees are bases, or indifferent 
compounds. This is the case with the oxides of manganese. 
Besides the three ah-eady mentioned, there are also 

Manganese. Oxygen. 

Manganic Acid 27.6 24 

Permangamc Acid 27.6 28 

The different oxides of nitrogen are, in like manner. 

Protoxide of Nitrogen 

Deutoxide of Nitrogen 

Nitro^<s Acid 

Hi/ponitroits. Acid 

Nitr^'c Acid 

It will be noticed that here, as m other places, the term 
oxides is used genericaUy for all the compounds of an element 
with oxygen, whether acids or bases, and the terms protoxide, 
&c. to designate the specific compounds. It is not unfre- 
quently the case that the same oxide acts as an acid under 
some circumstances, and as a base under others, and accord- 
ingly is known under two different names. "Water, when a 
base, is called Oxide of Hydrogen, but when an acid, it is 
named Hydric Acid. The oxide of aluminum, when a base, 
is called Sesquioxide of Aluminum, but when an acid, Alu- 
minic Acid. 

Next to the oxides, the compounds of sulphur with the 
elements are the most important binaries. These are called 
sulphides, and, as a general rule, for every oxide there is a 
corresponding sulphide, which is named after the analogy of 
the name of the oxide. Thus we have two sulphides of mer- 
cury : — 



NOMENCLATURE OF CHEMISTRY. 



/SMfeulphide of Mercury 
Protosulphide of Mercury. 

There ai'e three sulpliides of iron : - 

Protosulphide of Iron 
Sesquisulphide of Iron 
Persulphide of Iron 

The last corresponds to no oxide. The sulphides, like the 
oxides, may be divided into basic or indifferent sulphides, and 
into acid sulphides. The sulphur bases are named as above. 
The sulphur acids have been named by prefixing to the name 
of the oxygen acid the letters sulpho. Thus we have, cor- 
responding to 



[ercury. 
100 


Sulphur. 
8 


100 


16 


Iron. 
28 


Oxygen. 
8 


26 


12 


28 


16 



Arsenious Acid, 
Arsenic Acid, 
Carbonic Acid, 
Hydric Acid, 



SiilphoBXsemons Acid. 
Sulphoarsenic Acid. 
iS^M^^ocarbonic Acid. 
^M^Aohydric Acid. 



The names of the binary compounds of the other elements 
are named after the same analogy as the oxides and sulphides. 
Thus with the other elements 



Oxygen 

Fluorine 

Chlorine 

Bromine 

Iodine 

Sulphur 

Selenium 



forms 



Oxides. 

Fluorides. 

Chlorides. 

Bromides. 

Iodides. 

Sulj)hides. 

Selenides. 



Q NOMENCLATURE OF CHEMISTRY. 

Tellurium forms Tellurides. 

Nitrogen " Nitrides. 

Phosphorus " Phosphides. 

Arsenic " Arsenides, &c. 

"We distinguish the different fluorides, bromides, &c. by 
prefixes, as we do the oxides. Each of these classes of com- 
pounds may be subdivided, like the oxides or sulphides ; but 
only in a few cases has the nomenclature of the oxygen and 
sulphur acids been extended to them. Almost the only 
instances are the names of the compounds of the first few 
elements of the last list with hydrogen, which are some- 
times called Fluohydric, Chlorohydric, Bromohydric Acids, 
&c.', but even here the synonymous names Hydrofluoric, 
Hydrochloric, Hydrobromic Acids, &c. are more generally 
used. 

Names of Ternaries. — The name of a ternary compound, 
or salt, is very simply formed from the names of the acid and 
base of which it consists. The termination of the name of 
the acid is changed, if ic, into ate ; if ous, into ite ; and the 
name of the base added. Thus, sulphur^c acid and oxide of 
sodium form Sulphate of the Oxide of Sodium (or of Soda) ; 
sulphurot<s acid, Sulph^^e of the Oxide of Sodium. In like 
manner, hyposulphurzc acid forms Hyposulphafe, and hypo- 
sulphuroMS acid, Hyposulph^Ve of the Oxide of Sodium. So, 
also, Chlorite of Oxide of Potassium, and Chlorate of Oxide 
of Potassium ; Selenite of Oxide of Barium, and Selenate of 
Oxide of Barium ; Nitrite of Oxide of Calcium, and Nitrate of 
Oxide of Calcium ; Arsenite of Oxide of Lead, and Ai'senate 
of Oxide of Lead. When the base is an oxide of one of the 



NOMENCLATURE OF CHEMISTRY. 9 

heavy metals, the name of the salt is frequently abbreviated 
by leaving out the word oxide ; as, Sulphate of Lead, for Sul- 
phate of the Oxide of Lead ; and Nitrate of Copper, for 
Nitrate of the Oxide of Copper. When the base is an oxide 
of one of the light metals, by substituting for Oxide of So- 
dium, Oxide of Potassium, &c., the single words Soda, Po- 
tassa, &c., as already explained on page 3. Listead of Sul- 
phate of Oxide of Barium, chemists more frequently write 
the shorter Sulphate of Baryta; instead of Nitrate of Oxide 
of Calcium, Nitrate of Lime. "When the same acid combines 
with two or more different oxides of the same element to form 
salts, these are distinguished by introducing the specific name 
of the oxide iato the name of the salt ; as. Sulphate of Prot- 
oxide of Iron, and Sulphate of AS'es§'M^oxide of Iron ; Nitrate 
of ASwSoxide of Mercury, and Nitrate of Protoxide of Mer- 
cury. It is frequently the case that an acid combines with 
the same base in two or more different proportions, forming 
two or more different salts. In order to distinguish these 
salts, the one is called the neutral salt ; that containing more 
base than the neutral salt, a hasic ; and that containing less, an 
acid. Thus we have 

Basic Phosphate of Lime. 
Neutral Phosphate of Lime. 
Acid Phosphate of Lime. 



So, also, 



Basic Chromate of Lead. 
Neutral Chromate of Lead. 



When there are several basic salts, they are distinguished 
by Latin prefixes. Thus there are five acetates of lead : — 



10 NOMENCLATURE Or CHEMISTRT. 



Neutral Acetate of Lead 

Bibasic Acetate of Lead 

Sesquibasic Acetate of Lead 

Tribasic Acetate of Lead 

Sexbasic Acetate of Lead 
The acid salts are distinguished by placing the Latin prefixes 
directly before the name of the neutral salts ; as, 

Soda. Boracic Acid. 

Borate of Soda 31 35 

Biborate of Soda 31 70 



Oxide of Lead. 


Acetic Acid. 


112 


51 


224 


51 


280 


51 


336 


51 


672 


51 





Potassa. 


Chromic Acid. 


Neutral Chromate of Potassa 


47 


51 


Bichromate of Potassa 


47 


102 


Trichromate of Potassa 


47 


153 



Most of the oxygen acids are commonly used only when in 
combination with water. These compounds are true salts, in 
which the water plays the part of a base. The common 
Sulphuric Acid is a Sulphate of Water, and the common 
Nitric Acid, Nitrate of Water. Properly speaking, the terms 
Sulphuric and Nitric Acid ought only to be applied to the 
anhydrous compounds ; but custom authorizes us to extend 
these names to the hydrates. 

The names of the ternary sulphur compounds, the sulphur 
salts, are formed in the same way as those of the oxygen salts ; 
thus, the compound of sull^hocarbon^c acid and sulphide of 
sodium is called the Sulphocarbona^e of the Sulphide of So- 
dium ; the compound of sulphoarsen^c acid and sulphide of 
potassium, the Sulphoarsenafe of the Sulphide of Potassium ; 
and the compound of sulphoarsenioz<s acid and the same base, 
SulphoarsenzVe of the Sulphide of Potassium. 



NOMENCLATURE OF CHEMISTRY. 11 

The names of the ternary fluorine, chlorine, bromine, and 
iodine compounds are formed after a different analogy. These 
are generally regarded as double salts, and are called the 
double fluorides, chlorides, bromides, or iodides of the two 
metals which they contain. Hence the names 

Double Chloride of Aluminum and Sodium. 

Double Chloride of Platinum and Ammonium. 

Names of Quaternaries. — The double salts formed from 
two salts each contaiaing the same acid but different bases, 
are called double salts of the two bases. Thus, the compound 
of sulphate of alumina and sulphate of potassa is called the 
Double Sulphate of Alumina and Potassa ; the compound of 
sulphate of soda and sulphate of zinc, the Double Sulphate 
of Soda and Zinc. So, also, the Double Sulphate of Potassa 
and Magnesia, and the Double Sulphate of Potassa and Water. 
The last is more frequently called Bisulphate of Potassa, as if 
it were composed of one equivalent of base and two equiva- 
lents of acid, instead of being, as it probably is, a compound 
of two salts. Similar incongruities in the nomenclature, aris- 
ing frequently from different theories in regard to the consti- 
tution of substance, are not uncommon among the higher orders 
of compounds. 

The Lavoisierian nomenclature has not been found suffi- 
ciently expansive to meet the requirements of modern sci- 
ence, not only on account of the complex character of many 
of the newly discovered compounds, but more especially 
because it involves the ideas of a peculiar theory, which, 
although once almost universally received, is not now so gen- 
erally admitted. The inadequacy of the old nomenclature to 



12 NOMENCLATUKE OF CHEMISTRY. 

afford names for the great variety of chemical compounds, or 
to describe the infinite changes to which they are hable, has 
given rise to a peculiar chemical language, analogous to the 
language of mathematics, called Chemical Symbols. To ex- 
plain the use of this language will be the object of the next 
chapter. 



CHEMICAL SYMBOLS. 



Since all matter is composed of one or more of sixty-two 
different substances, never as yet decomposed, and hence called 
Elements, it is evident that, if we adopt an arbitrary symbol for 
each of these elements, we shall be able, by combining them to- 
gether, to express aU possible varieties of combination. More- 
over, since the elements always combine in certain fixed and 
definite proportions by weight, it is equally evident that, if we 
assign to each of these symbols a certain weight, we shall be 
able to indicate the relative quantities of the difierent elements 
which enter into any compound. 

Symbols of Elements. — It has been agreed by chemists of 
different nations to use, as symbols of the elements, the first 
letters of their Latin names. "When two or more names com- 
mence with the same letter, a second letter is added for dis- 
tinction. The first letter is printed or written in capitals, and 
the second, when used, in small letters, immediately following 
the first. A list of the Elements, with their Symbols, is given 
on the following page. 
2 



14 



CHEMICAL SYMBOLS. 



CHE]\nCAL SYIVEBOLS A2TD EQUIVALENTS. 



Aluminum 


Al = 


13.7 


Xickel 


Ni = 


29.6 


Antimony (Stibium) 


Sb = 


129 


^Niobium 


Nb 




Arsenic 


As = 


75 


Nitrogen 


N = 


14 


Barium 


Ba = 


68.5 


Xorium 


No 




Bismuth 


Bi = 


213 


Osmium 


03 = 


99.6 


Boron 


B = 


10.9 


Oxygen 


= 


8 


Bromine 


Br = 


80 


Palladium 


Pd = 


53.3 


Cadmium 


Cd = 


56 


Pelopium 


Pe 




Calcium 


Ca = 


20 


PJios2)horus 


P = 


32 


Carbon 


C = 


6 


Platinum 


Pt = 


98.7 


Cerium 


Ce = 


47 


Potassium (Kalium) 


K = 


39.2 


Chlorine 


CI = 


35.5 


Rhodium 


E = 


52.2 


Chromium 


Cr = 


26.7 


Euthenium 


Eu = 


52.2 


Cobalt 


Co = 


29.5 


vSelenium 


Se = 


89.5 


Cop/>er (Cuprum) 


Cu = 


31.7 


Silicon 


Si = 


21.3 


Didymium 


D 




Silver (Argentum) 


Ag = 


lOS.l 


Erbium 


E 




Sodium (Katrimn) 


Na = 


23 


Fluorine 


Fl = 


18.9 


Strontium 


Sr = 


43.8 


Glucinum 


G = 


4.7 


Sulphur 


S = 


16 


Gold (Aurum) 


Au = 


197 


Tantalum 


Ta = 


184 


Eydrogen 


H = 


1 


Tellurium 


Te = 


64.2 


Iodine 


I = 


127.1 


Terbium 


Tb 




Iridium 


Ir = 


99 


Thorium 


Th = 


59.6 


Iron (Ferrum) 


Fe = 


2S 


Tin ( Stannum) 


Sn = 


59 


Lanthanium 


La 




Titanium 


Ti = 


25 


Lead (Plumbum) 


Pb = 


103.7 


Tungsten ( 'Wolfram) 


W = 


95 


Lithium 


Li = 


6.5 


Uranium 


U = 


60 


Magnesium 


Mg = 


12.2 


Vanadium 


V = 


68.6 


Manganese 


Mn = 


27.6 


Yttrium 


Y 




Mercury (Hydrargyrum^ 


Hg = 


100 


Zinc 


Zn = 


32.6 


Molybdenum 


Mo = 


46 


Zirconium 


Zr = 


22.4 



CHEMICAL SYMBOLS. 15 

The student will do well to notice in the foregoing list the 
symbols of those elements whose Latin names commence with 
letters differing from the initial letters of the English names, 
since they are not so easily remembered as the others. 

Chemical Equivalents. — The chemical symbols not only 
stand for the names of the elements, but also for a fixed pro- 
portional weight of each. These weights are given in the 
above table, opposite to the symbols. They have only rela- 
tive values ; if one is in pounds, all the rest are in pounds ; 
and if one is in ounces, all the rest are in ounces. We may 
leave the standard indefinite, and express the weight in parts ; 
then Al stands for 13.7 parts of aluminum ; Sb stands for 129 
parts of antimony, &c. The weight of an element indicated 
by its symbol is called one equivalent, and it is a law of 
chemistry that elements always combine by equivalents ; that 
is, one equivalent of one combines "with one equivalent of an- 
other, or else several equivalents of one combine with one or 
with several equivalents of another. 

As stands for 75 parts of Arsenic, or one equivalent. 
BK " 68.5 " Barium, « " 

H " 1 " Hydrogen, « « 

" 8 « Oxygen, " « 

In order to express two, three, or more equivalents of an 
element, we place a figure just below the symbol at its right 
hand; thus, 

02 stands for 16 parts of Oxygen, or two equivalents. 

03 " 24 " " three equivalents. 

These figures merely multiply the symbols beneath which 



16 CHEMICAL SYMBOLS. 

they stand, and must not be confounded with algebraic powers, 
which are sometimes written in a similar way. We sometimes 
place the figure, though larger, before the symbol ; 2 means 
exactly the same thing as Oj. 

Symbols of Compounds. — In order to form the symbol of a 
compound, Ave write the symbols of the elements of which it 
consists one after the other, indicating by means of figures the 
number of equivalents of each which have entered into com- 
bination. Thus, H is the symbol of water, a compound con- 
sisting of one equivalent or one part of hydrogen, and of one 
equivalent or eight parts of oxygen ; S O3 is the symbol of 
sulphuric acid, a compound consisting of one equivalent or six- 
teen parts of sulphur, and of tliree equivalents or twenty-four 
parts of oxygen ; C12 Hn On is the s^bol of common sugar, 
a compound consisting of twelve equivalents of carbon, eleven 
equivalents of hydrogen, and eleven equivalents of oxygen. 

Binary Compounds. — The symbols of binary compounds 
are formed by writing the symbols of the two elements to- 
gether, taking care to place the symbol of the metal, or of 
the most electro-positive element, first. The binary symbol thus 
obtained represents always one equivalent of the compound. 
The weight of this equivalent is evidently the sum of the 
weights of the equivalents of the elements entering into the 
compound. 

14 + 40 = 54. 

N O5 stands for one equiv. or 54 parts of Nitric Acid. 

16 + 24 = 40. 

S O3 " " 40 « Sulphuric Acid. 



CHEMICAL SYMBOLS. 17 

6 + 16 = 22. 

C O2 stands for one equiv. or 22 parts of Carbonic Acid. 

1 + 8 := 9. 

II « « 9 « Water. 

28 + 8 = 36. 

FeO " « 36 " Oxide of Iron. 

20 + 8 = 28. 

CaO " " 28 « Lime. 

23 + 8 = 31. 

Na " " 31 « Oxide of Sodium. 

6 + 32 = 38. 

CS2 " " 38 " Sulphocarbonic Acid. 

75+ 48 = 123. 

As S3 " 123 " Sulphoarsenious Acid. 

28 + 16 = 44. 

FeS " " 44 « Sulphide of Iron. 

39 + 16 = 55. 

KS " " 55 " Sulphide of Potassium. 

In order to express two or more equivalents of a binary, we 
place a figure immediately before the symbol, like an algebraic 
coefficient. A figure so placed always multiplies the whole 
hinary. 

3 S O3 stands for 3 equiv. or 120 parts of Sulphuric Acid. 
5PbO " 5 " 560 « Oxide of Lead. 

Ternary Compounds. — The symbol of a ternary compound 
is formed by writing together the symbols of the two binaries 
of which it consists, separated by a comma, taking care to 
place the most electro-positive binary, the base, first. If the 
salt" is composed of more than one equivalent of either base or 
acid, then the number of equivalents must be indicated by 
2* 



18 CHEMICAL SYMBOLS. 

coefficients. The ternary symbol thus obtained always stands 
for one equivalent of the compound, and the weight of this 
equivalent is evidently the sum of the weights of the equiva- 
lents of the elements entering into it. 

1 + 8 + 16 + 24 = 49. 

H O, S O3 stands for 1 equiv. or 49 parts of Sulphate of Water 
(common Sulphuric Acid). 

1 + 8 + 14 4- 40 = 63. 

H O, N O5 stands for 1 equiv. or 63 parts of Nitrate of Water 
(common Nitric Acid). 

39 + 8 + 6 + 16 = 69. 

K O, C O2 stands for 1 equiv. or 69 parts of Carbonate of 
Potassa. 

23 + 8 + 16 -(- 24 = 71. 

Na 0, S O3 stands for 1 equiv. or 71 parts of Sulphate of Soda. 

39 + 8 + 27 + 24 = 98. 

K O, Cr O3 stands for 1 equiv. or 98 parts of Neutral Chromate 
of Potassa. 

39 + 8 + 2 (27 + 24) = 149. 

K 0, 2 Cr O3 stands for 1 equiv. or 149 parts of Bichromate of 
Potassa. 

39 + 8 + 3 (27 + 24) = 200. 

K 0, 3 Cr O3 stands for 1 equiv. or 200 parts of Trichromate of 
Potassa. 

104 + 8 4- 27 4- 24 = 163. 

PbO, CrOs stands for 1 equiv. or 163 parts of Neutral Chro- 
mate of Lead. 

2 (104 + 8) + 27 4- 24 = 275. 

2 PbO, Cr03 stands for 1 equiv. or 275 parts of Basic Chromate 
of Lead. 

In order to express two or more equivalents of a ternary, 
we enclose the symbol in parentheses, and place before the 
whole the required figure. Thus, 



CHEMICAL SYMBOLS. 19 

3 (Na 0, S Og) stands for three equivalents of Sulphate of Soda. 
5 (2 Pb O, Cr O3) stands for five equivalents of Basic Chromate 
of Lead. 

Neutral Salts. — The larger number of inorganic acids com- 
bine most readily with one equivalent of base, and the salts so 
formed will be called neutral salts. If the salts contain more 
equivalents of acid or base than one, they are called acid or 
basic salts respectively. There are, however, some acids which 
combine most readily with two or three equivalents of base, in 
the same way that the others combine with one. Such acids 
are called bibasic or tribasic acids, in order to distinguish them 
from the rest, which are frequently called monobasic. Of 
bibasic and tribasic acids, the most important in inorganic 
chemistry is Phosphoric Acid. This Is known in three differ- 
ent conditions. In the first of these it is monobasic, in the 
second bibasic, and in the third tribasic, the last being the 
ordinary condition. The three conditions are designated by 
the symbol aP O5 ; ^P O5 ; ^P O5. The acid eP O5 forms neutral 
salts when combined with three equivalents of base, the acid 
bP Og when combined with two, and the acid ^P O5 when com- 
bined with only one. There are three compounds of the acid 
and water corresponding to the three conditions, which are rep- 
resented in symbols by H O, ^P Oj ; 2 H 0, tP Og ; 3 H O, ,P O5. 
In these compounds we can substitute for the equivalents of 
water equivalents of other bases, either in whole or in part, 
forming such compounds as Na O, ^P Og ; 2 Na O, bP Og ; 
3]S-aO,,P05 ; [HO, 2 NaO] ^P O5 ; [2 H 0, Na 0] ^P Og. 
The equivalents of water may even be replaced by different 
bases, as in the compounds [Na O, Pb 0] bP Og ; [H 0, Na 0, 



20 CHEMICAL STJIBOLS. 

K 0], eP O5 ; [K 0, 2 Mg 0] ,T O5. All the above are sym- 
bols of neutral salts. 

As protoxide bases combine most readily with one equivalent 
of acid, so sesquioxide bases combine most readily mth three 
equivalents. A neutral salt of a sesquioxide base is therefore 
one which contains for every equivalent of base three equiva- 
lents of a monobasic acid, or one equivalent of a tribasic acid, 
and for every two equivalents of base three equivalents of a 
bibasic acid. Hence, Fca O3 , 3 S O3 ; 2 Fcg O3 , 3 tP O5 ; 
P62 O3 , cP O5 are all symbols of neutral salts. On the other 
hand, Fea O3 , S O3 ; 4 AI2 O3 , 3 ,P O5 ; 2 Al^ O3 , cP O5 are sym- 
bols of basic salts. It will be noticed, on examining the above 
symbols, that neutral salts of monobasic acids contain as many 
equivalents of acid as there are equivalents of oxygen in the 
base, and neutral salts of bibasic and tribasic acids one half 
and one third as many, respectively. This rule must be kept 
in mind when writing the symbols of salts. 

Compound Radicals. — There is a large class of substances 
which, although compound, nevertheless act in chemical changes 
exactly as if they were simple, frequently replacing the ele- 
ments themselves. Such substances are termed compound 
radicals. Many of these radicals have, like the elements, 
received arbitrary names, such as Cyanogen, Ammonium, 
Ethyle, Acetyle, &c. ; and, moreover, the first letter or letters 
of these names are frequently used as their symbols. It is 
best, however, to write out the symbols of the elements form- 
ing these compounds, and enclose the whole in brackets ; thus, 
[N H4] stands for Ammonium, [Cj N] for Cyanogen, [C4 H5] 
for Ethyle, [C4 H3] for Acetyle. The oxides of the compound 



CHEMICAL SYMBOLS. 21 

radicals, like those of the elements, may be divided into acids 
and bases, and their symbols are Avritten exactly like those of 
other binaries. Thus, 

[C4 H3] O3 is the symbol of Acetic Acid. 

[N H4] O " " Oxide of Ammonium. 

[C4 H5] " " Oxide of Ethyle. 

So, also, with the salts. 

[N H4] O, [C4 H3] O3 is the symbol of Acetate of Ammonia. 

[C4 Hs] 0, [C4 H3] O3 " " Acetate ofOxide of Ethyle. 

Water of Crystallization. — Besides the water of constitution, 
which frequently forms a part or the whole of the base of a 
salt, most salts combine with water as a whole. This water is 
held in combination by a comparatively feeble affinity, and 
may be generally driven off by exposing the salt to the tem- 
perature of 100° C, and sometimes escapes at the ordinary 
temperature of the air, the crystals of the salt in all cases 
fallmg into powder. Its presence is essential to the crystalhne 
condition of many salts, and hence the name "Water of Crys- 
tallization. The presence of water of crystallization in a salt 
is expressed in symbols, by writing after the symbol of the 
salt, and separated from it by a period, the number of equiva- 
lents of water. Thus, 

Fe O, S O3 . 7 H is the symbol of Crystallized Sulphate of 

the Oxide of Iron (Green Vitriol). 
H 0, 2 Na O, eP O5 . 24 H O is the symbol of Crystallized 

Phosphate of Soda. 

The same salt, when crystallized at different temperatures, not 

unfrequently combines with different amounts of water of crys- 

, tallization, the less amounts corresponding to the higher tem- 



22 CHEMICAL SraiBOLS. 

peratures. Thus, the Sulphate of Manganese may be crystal- 
lized with three different amounts of water of crystallization. 

Mn O, S O3 . 7 H O when crystallized below 6° Centigrade. 
Mn 0, S O3 . 5 H O " « between 7° and 20°. 
Mn O, S O3 . 4 H O « « between 20° and 30°. 

The crystalline forms of these three compounds are entirely 
different from each other, proving that the form depends, in 
part at least, on the amount of water which the salt contains. 

The symbols of other ternary compounds are written like 
those of the oxygen salts, and therefore require no further 
explanation. Below are a few of these symbols, together with 
those of the corresponding oxygen salts, which may serve as 
examples. 

K O, C O2 = Carbonate of Oxide of Potassium. 

K S, C S2 = Sulphocarbonate of Sulphide of Potassium. 

K O, As O3 = Arsenite of Oxide of Potassium. 

K S, As S3 = Sulphoarsenite of Sulphide of Potassium. 

3 Na CI, Sb CI3 = Double Chloride of Antimony and Sodium. 
[NH4] CljPtCla = Double Chloride of Platinum and Potassium. 
K I, Pt I2 = Double Iodide of Platinum and Potassium. 

Quaternaries. — The symbols of the double salts are formed 
by writing together the symbols of the two salts of which they 
consist, separated by a period. Thus, 

K O, S O3 . iMg O, S O3 . 6 H O = Double Sulphate of Mag- 
nesia and Potassa. 

K O, S O3 . AI2 O3, 3 S O3 . 24 H == Double Sulphate of Alu- 
mina and Potassa (Alum). 

When the salts contain water of crystallization, the amount of 
this water expressed in equivalents is written after the symbol 
of the salt, as already explained. 



CHEMICAL EEACTIONS. 



The various chemical changes to which all matter is more 
or less liable are termed, in the language of chemistiy, reac- 
tions, and the agents which cause these changes, reagents. In 
every chemical reaction we must distinguish between the sub- 
stances which are involved in the change and those which are 
produced by it. The first will be termed the factors, and the 
last the products, of the reaction. As matter is indestructible, 
it follows that The sum of the weights of the products of any 
reaction must always be equal to the sum of the weights of the 
factors. This statement seems at first sight to be contradicted by 
experience, since wood and many other combustible substances 
are apparently consumed by burning. In all such cases, how- 
ever, the apparent annihilation of the substance arises from 
the fact that the products of the change are invisible gases ; 
and when these are collected, their weight is found to be equal, 
not only to that of the substance, but also, in addition, to the 
weight of the oxygen from the air consumed in the process. 
As the products and factors of every chemical change must be 
equal, it follows that A chemical reaction may always be repre- 



24 CHEMICAL REACTIONS. 

sented in an equation hy writing the symbols of the factors in 
the first member, and those of the products in the second. The 
I'eaction of sulphuric acid on common salt may be represented 
by the following equation : 

23 + 35 + 1+8 + 16+24 = 23 + 8 + 16+24 + 1 + 35 = 107. 

NaCl + ^0, SOs = Mi 0,SOs +HC1. 

The correctness of this may be proved by adding together the 
equivalents of both sides, when the sums will be found to be 
equal. In like manner, the reaction of a solution of common 
phosphate of soda on a solution of chloride of calcium may be 
represented by the equation 

1+8 + 2 (23 + 8)+ 32+40 + 3(20+35) = 

IIO,2I^aO, ,FOs-^SCaCl-\-Aq* = 

3 (20 + 8) + 32+40 + 2(23 + 35) + 1 + 85 = 308. 

3CaO, eP05 + 2iVa CI -\- JI CI -\- Aq. 

So, also, the reaction of hydrochloric acid on chalk, which may 
be proved like the other two : 

CaO,CO^-{- !£ CI -\-Aq= CaCl-]-II0-{-Aq-\-COa. 

Although the equation is the most concise, and therefore 
in most cases the best form of representing chemical reac- 
tions, it is nevertheless frequently advantageous, in studying 
comphcated changes, to adopt a more graphic method, by 
which the various steps of the process may be indicated. The 
reactions represented by the preceding equations may be writ- 
ten thus : — 



* The symbol Aq, for Aqua, merely indicates the condition of solution, and 
is not to be regarded in adding up the equivalents in order to prove the equa- 
tion. 



CHEMICAL REACTIONS. 



25 




(2.) 



HO,2XaO,„ PO^ 




Chemical reactions may be classed under three divisions. 

First, those reactions in which a compound is decomposed, 
and divides into simpler compounds or into elements. E. g. 
when oxide of mercury is heated, it is decomposed into oxygen 
gas and metallic mercury. Thus, 

Hg0 = ^y + O. 

Again, when Chlorate of Potassa is heated, it is resolved into 
oxygen gas and chloride of potassium. Thus, 

K O, CI O5 == K CI + 6 O. 
3 



26 CHEMICAL REACTIONS. 

So, also, when sulphate of lead is heated, it is resolved into 
anhydrous sulphuric acid and oxide of lead. Thus, 

PbO,S03= PbO+SOs. 

Such reactions as these Avill be called analytical, and the 
process analysis. 

Second, those reactions in which the elements are united to 
form compounds, or compounds of a lower order to form those 
of a higher. E. g. when hydrogen and carbon bum in the air, 
they combine with oxygen to form water or carbonic acid. 
Thus, 

+ 0=HO; C + 02=C02. 

Again, when anhydrous sulphuric acid combines with lime to 
form sulphate of lime. Thus, 

CaO + S03=CaO,S03. 

Reactions like these will be called sy7ithetical, and the process 
synthesis. 

Third, those reactions in which one element displaces an- 
other. E. G. when sodium takes the place of hydrogen in 
water, or zinc the place of hydrogen in dilute sulphuric acid. 
Thus, 

^6> + Na = iVa(9 + H. 
Zn-^HO,SO^-\-Aq=ZHO,SO,-\-Aq^U.. 

This division includes also those reactions in which there is 
a mutual interchange of elements between two compounds. 
E. g. when a solution of chloride of barium is added to a 
solution of sulphate of soda, the sodium and barium change 
places, and we have formed an insoluble precipitate of sulphate 



CHEMICAL REACTIONS. 



27 



of baryta and chloride of sodium (common salt), which remains 
in solution. Thus, 

Ba a-\-Na 0, S 0^-^- Aq = Bs, 0,^0, -\- Fa Cl+Aq. 

Reactions like these will be called metathetical, and the process 
metathesis* 

Of the three classes of chemical reactions, the last is by far 
the most important ; indeed, the larger number of reactions 
described in an elementary treatise on chemistry are examples 
of metathesis. All metathetical reactions can be illustrated 
very elegantly with the aid of mechanical diagrams, as fol- 
lows : — 

(4) (5.) 




They are easily made by printing with stencils on the larger 
piece of pasteboard, A B, Fig. 4, the symbols of the elements 
not disturbed in the reaction, and on the smaller piece, a b, the 
symbols of the elements which exchange places. The smaller 
piece having been fastened to the larger by means of an eye- 
let at O, the reaction is represented by merely turning it 
half round. (See Fig. 5.) If the symbols of the interchang- 
ing elements are not symmetrical on all sides, it is of course 
necessary to make them reversible, by printing each on a sep- 
arate small square of pasteboard, fastened by an eyelet to the 



* From the Greek /aerar I'^Tj/xt, to displace or to transpose. 



28 CHEMICAL REACTIONS. 

top or bottom of the revolving piece a b, since otherwise the 
letters would be inverted when the diagram is turned. This 
method of illustration may, with a little ingenuity, be extended 
to some of the most complicated cases of chemical change. 

The most important condition of chemical action is, that the 
particles of the substances involved in the change should be 
indued with freedom of motion. This condition is generally 
fulfilled, both in nature and in our laboratories, by bringing the 
substances together in solution, either in water or in some other 
fluid. When substances are brought together in solution, there 
are two circumstances which, more than any others, determine 
the nature and extent of the resulting change. 

First, If hy an interchange of analogous elements an insoluble 
coynpound may he formed, this compound always separates from 
the fluid as a precipitate. As this circumstance is by far the 
most important of all in determining chemical reactions, it 
requires full illustration. 



Ba 0, N 0: 
HO, SO, 



1 , . HO,NOs\ . 
|+^^=Ba6,S0:|-^?- 



In these examples, and in general throughout the volume, the 
symbols of the substances, when in solution, are printed in 
italic letters, and the solid precipitate in Roman letters. The 
symbol Aq, as already stated, stands for an indefinite amount 
of water, in which the substances are supposed to be dissolved. 
It is obvious, from the above examples, that, in order to ascer- 



CHEMICAL REACTIONS. 29 

tain whether two salts will react on each other, when brought 
together in solution, so as to form a precipitate, it is only neces- 
sary to Avrite the symbol of one under that of the other, and 
interchange the symbols of the metallic elements. If either 
compound whose symbols are thus formed is insoluble in the 
menstruum present, a reaction will take place, and the insoluble 
compound will be precipitated. At the end of the volume will 
be found a table, reprinted from the English edition of Fre- 
senius's Qualitative Analysis, by means of which the student 
can easily ascertain from inspection the solubility of any of 
the more frequently occurring binary compounds or salts, and 
thus will be able to solve the following problems. 

Problem 1. If chloride of barium and sulphate of soda are 
mixed together in solution, wiU there be a reaction ; and if so, 
what will be formed ? 

Problem 2. If chloride of sodium and nitrate of silver are 
mixed together in solution, will there be a reaction, &c. ? 

Problem 3. If sulphide of hydrogen and nitrate of lead are 
mixed together in solution, will there be a reaction ? 

Problem 4. If sulphide of hydrogen and sulphate of zinc 
are mixed together in solution, will there be a reaction ? 

In solving the last problem, it must be noticed that the fluid 
which would result from a reaction would be a weak acid, in 
which many substances are soluble which would be insoluble 
in pure water, as may be seen from the table. 

Problem 5. If chloride of sodium and sulphate of copper 
are mixed together in solution, will there be a reaction ? 

Problem 6. If sulphuric acid and borate of soda are mixed 
together in solution, will there be a reaction ? 

It will be found that, by an interchange of metallic elements 
3* 



30 CHEMICAL REACTIONS. 

in the last two examples, no insoluble compound will be formed, 
and hence the conclusion follows from our data, that there will 
be no precipitate. We must not, however, conclude from this 
that there will be no reaction, since, as can easily be seen, it 
does not necessarily follow, because the possible formation of an 
insoluble compound always determines a reaction, that the re- 
verse is equally true, and that no reaction can take place unless 
an insoluble compound is formed. Indeed, in the last two ex- 
amples, we are able, from incidental phenomena, to determine 
satisfactorily that a change does result ; thus, in Problem 5, 
when the solutions are mixed, the blue color of sulphate of 
copper changes into the green color of chloride of copper ; 
and in Problem 6, if sulphuric acid is not added in excess, the 
claret color to which blue litmus-paper turns in the mixed 
solution proves that it is boracic acid, and not sulphuric acid, 
wlaich is in a free state ; nevertheless, in most similar cases it 
is impossible to determine, unless an insoluble compound is 
formed, whether any reaction has taken place. 

Second. The circumstance which, next to insolubility, is 
most important in determining metathetical reactions, is vola- 
tility, and it may be laid down as a general principle, that, If 
hy an interchange of analogous elements between two substances 
in solution, a substance can be formed, which is volatile at the 
temperature at which the experiment is conducted, such an inter- 
change cdioays takes place, and the volatile product is set free. 
In order to illustrate this principle, a few examples may be 
adduced. 

1. If diluted sulphuric acid is poured upon granulated zinc, 
a brisk evolution of hydrogen gas ensues, and sulphate of oxide 
of zinc is retained in solution. Thus, 



CHEMICAL REACTIONS. 31 

Zn 1 jH 

HO,SO^-{-Aq\ — \ZnO,SO,-\-Aq. 

In this example, and those that follow, the volatile or gaseous 
products are always printed with a full-face type. 

2. If diluted sulphuric acid is poured upon protosulphide of 
iron, sulphide of hydrogen gas escapes, and sulphate of protox- 
ide of iron remains in solution. Thus, 

FeS ) (HS 

HO,SO^-{-Aq] — \Fe 0,SO^-\-Aq. 

It will be noticed from the last two reactions, that it is not 
essential that more than one of the factors of the reaction 
should be fluid, or in solution. 

8. If strong sulphuric acid is poured upon common salt, and 
the mixture slightly heated, chlorohydric acid gas is evolved, 
and bisulphate of soda remains dissolved in the excess of sul- 
phuric acid. Thus, 

Na a 7 f H CI 



} = 



HO,SOs . HO, ^6>3 1 — XNa 0, S O3 . HO, S 0,. 

4. If strong sulphuric acid is poured upon nitre, and the 
temperature of the mixture slightly elevated, the vapor of 
nitric acid is given off, and bisulphate of potassa is formed. 
Thus, 

KO,NO, \ fH0,JV05 

HO, 8 0^.H0,S0^\ — \K0, S O3 . HO, S O3. 

5. If diluted nitric acid is poured upon chalk, or any other 
analogous carbonate, carbonic acid gas is set free, and a salt 
of nitric acid formed. Thus, 

CaO, CO, ) _ j H0-\-CO, 

HO,NO,^Aq\ — \ OaO,J^Os-\-Aq. 



32 CHEMICAL KEACTIONS. 

It is very frequently the case that two, or even all of the 
three, classes of chemical reactions are combined, and going on 
simultaneously, in a single experiment. In the last example, 
for instance, the metathesis is succeeded by an analysis of one 
of the products, owing to the want of aflBnity between C O2 and 
H O. Again, the reaction of nitric acid on copper is an ex- 
ample where all three varieties of reactions are combined. 
Three equivalents of copper react on four equivalents of nitric 
acid, and the reaction may be conveniently studied in two parts. 
In the first part, the copper is oxidized by one of the equiva- 
lents of the acid ; here Ave have analysis accompanied by syn- 
thesis ; and in the second part the copper changes place with 
the hydrogen of the acid, a case of metathesis. 

1st. 3Cu + ^0, iV^(95 = 3(CuO, H0)-f-]\O2. 

HO,NO,\, SGuO,NOs 1 , . 

Z"^- CuO, H0[ +^^~ l^O + ^Oj +^^- 

The whole reaction combined may be expressed thus : 

3 Cu + 4 {HO, NO,) -{-Aq = 3(at 0,N0,) +^g + JVO^. 

Such mixed processes are very common in all complex cases 
of chemical change. 

StocJtiometrieal Problems. — The chemical symbols enable 
us not only to represent chemical changes, but also to calculate 
exactly the amounts of the substances required in any given 
process, as well as the amounts of the products wliich it will 
yield. The method of making such calculation can best be 
illustrated by examples. In these examples the weights and 
measures will be given according to the French decimal system, 
which is now very generally used in chemical laboratories, and 



CHEMICAL REACTIONS. 33 

which, on account of the very simple relation between the units 
of measure and of weight, greatly facilitates stochiometrical 
calculations. The French measures and weights can, when 
required, be very easily reduced to the English standards, by 
means of a table at the end of the volume. 

Pi'oblem 1. We have given 10 kilogrammes of common 
salt, and it is required to calculate how much chlorohydric acid 
gas can be obtained from it by treating with sulphuric acid. 
The reaction is 

23+35.5 = 58.5. 1+35.5 = 36.5. 

I^eiCl-{- ffO,S03-{-Aq = mO,SOs-\-Aq-\-U€L 

Hence one equivalent, or 58.5 parts, of common salt, will yield 
one equivalent, or 36.5 parts, of chlorohydric acid gas. There- 
fore the amount which 10 kilogrammes will yield can be cal- 
culated from the proportion 

53.5 36.5 

Na CI : H CI = 10 : X = 6.239 kilogrammes, Ans. 

Problem 2. It is required to calculate how much chlorine 
gas can be obtained from chlorohydi'ic acid with 6 grammes of 
hyperoxide of manganese. The equation representing the 
reaction is 

28.6+16 44.6. 35.5. 

Mn02 -\- 2 H 01 -\- Aq = Ml a -\- 2 HOi^Aq + €1. 

Hence the amount of hyperoxide of manganese represented by 
Mn O2, or 44.6 parts, yields an amount of chlorine gas repre- 
sented by CI, or 35.5 parts, and we have the proportion, 

44.6 35.5 
Mn O2 : CI = 6 : a; = 4.775 grammes. 

Problem 3. It is required to calculate how much sulphuric 



34 CHEBIICAL REACTIONS. 

acid and nitre must be used to make 250 gi'ammes of the 
strongest nitric acid. 

101.2 98 63 

K0,-^0,+2(JI0,S0,)=K0,S0s.II0,S0s-\-nO,NO,. 
Hence, 

63 98 

H0,I^0si2{II0,S O3) = 250 : x. 

63 101.2 

HO,NO^ : K 0, N Og = 250 : x. 

From the above examples we can deduce the following gen- 
eral rule for calculating from the amount of any given factor 
of a chemical reaction the amount of the products, or the 
reverse. Express the reaction in an equation : make then the 
proportion, As the symbol of the substance given is to the sym- 
bol of the substance required, so is the amount of the substance 
given to x, the amount of the substance required ; reduce the 
symbols to numbers, and calcidate the value of x. 

On account of the very great lightness, the amount of a gas 
is very much more frequently estimated by measure than by 
weight. At the end of the volume a table will be found giving 
the weight in grammes of one thousand cubic centimetres, or 
one litre, of each of the most important gases. With the aid 
of this table such problems as the following may be solved. 

Problem 4. How much chlorate of potassa must be used to 
obtain one litre of oxygen gas? One litre of oxygen gas 
weighs 1.43 grammes. 

K 0, CI O5 = K CI + 6 O. 

48 122.7 

6 : K 0, CI O5 = 1.43 : x. 



CHEMICAL REACTIONS. 35 

Problem 5. How much zinc and sulphuric acid must be used 
to obtain 4 litres of hydrogen gas ? Four litres of hydrogen 
weigh 0.357 grammes. 

Zn -^HO,SOs-\-Aq= Zn 0, S 0^ -\- Aq -\- H. 

1 32.6 

H : Zn == 0.357 : a; = amount of zinc required. 

1 49 

H : HO, S03 = 0.357 : x = amount of sulphuric acid required. 

Problem 6. If ten grammes of water are decomposed by 
galvanism, how large a volume of mixed gases will they give ? 

^0 = M+O. 

9 1 

JIO : H = 10 : x = i^- grammes of hydrogen.^ 

9 8 

^0:0=10 : x = ^^- grammes of oxygen. 

J^ grammes of hydrogen occupy 12.^429 cubic centimetres. 
-^ grammes of oxygen occupy 6.214 " " 

The mixed gases occupy 18,644 " " 

And hence * water, when decomposed into its elements, expands 
1864 times. 

The use of chemical symbols, both in expressing chemical 
reactions and in stochiometrical calculations, having been ex- 
plained, they will be used in the following pages to illustrate 
the text of Stockhardt's Elements of Chemistry. The reac- 
tions described in that work are represented in the form of 
equations, the first member containing always the factors, and 

* It must be remembered that one cubic centimetre of water weighs one 
sramme. 



36 CHEMICAL KE ACTIONS. 

the second, the products of the process. It remams for the 
student to work out the reaction, and represent it in the manner 
explained on page 25. The equivalents of water which are set 
free or formed during a reaction, are not generally indicated, 
but are merged in the general symbol ^5'; and the student 
will frequently be obliged to supply these equivalents in order 
to work out the reaction. The symbols of solids are printed 
in Roman letters, those of fluids in italics, and those of gases 
in full-faced type. "When, however, the solid or gas is dis- 
solved in water, the symbol is printed in italics, followed by 
the general symbol Aq, or aq. Aq always stands for an indefi- 
nite and large amount of water ; ag, for an indefinite but small 
amount of water. The color of a substance, especially of pre- 
cipitates, is frequently printed above its symbol. The head- 
ings and figures, or letters at the side of the page, refer to the 
sections of the above-mentioned book. 

In order still further to illustrate the subject, a large number 
of problems have been added to the reactions, which the stu- 
dent is expected to solve. The method of solving these prob- 
lems can, in most cases, be deduced from the explanations 
already given, and from the sections of the text-book ; in all 
other cases the method is explained under the problem. 
Throughout the following pages, all gases and vapors are sup- 
posed to be measured at the temperature of 0° and when the 
barometer stands at 76 centimetres, unless some other tempera- 
ture or barometric pressure is expressly stated. All specific 
gravities are referred to water at its maximum density (at 4°). 
The temperatures are all given on the Centigrade scale, and 
the weights and measures according to the French decimal 
system. 



EEACTIONS AND PROBLEMS. 



WEIGHING AND MEASUEING. 



10. Problems on French System of Weights and Measures. 

1. Reduce by means of the table at the end of the book, — 
a. 30 inches to fractions of a metre. 

h. 76 centimetres to English inches. 

c. 36 feet to metres. 

d. 10 metres to feet and inches. 

2. Reduce by means of the table at the end of the book, — 
a. 8 lbs. 6 oz. to grammes. 

h. 7640 grammes to English apothecaries' weight. 
c. 45 grains to grammes. 

3. "What is the diameter, and what is the circumference, of 
the globe in French measure ? 

4. What is the distance from Dunkirk in France to Barce- 
lona in Spain ? The latitude of Dunkirk =51° 3', that of 
Barcelona = 41° 22', and the two places are on the same 
meridian. 

5. "Were our globe composed entirely of water at its great- 
est density, what would be its weight in kilogrammes ? 

6. What is the weight of one cubic decimetre of water ? 

7. Reduce by means of the table at the end of the book, — 
a. 4 pints to litres and cubic centimetres. 

h. 5 gallons to litres and cubic centimetres. 



40 WEIGHING AND MEASURING. 

c. 5 litres to English measure. 

d. 4 cubic centimetres to English measure. 

11-17. Problems on Specijlc Gravity. 

1. The specific gravity of iron = 7.84. What is the weight 
of 1 cubic centimetre, 4 cubic centimetres, &c. of the metal m 
grammes ? 

2. The specific gravity of alcohol = 0.81. "What is the 
weiglit of one litre in grammes ? of 45 cubic centimetres, &c. ? 

3. The specific gravity of sulphuric acid = 1.85. K you 
wish to use in a chemical experiment 250 grammes, how much 
must you measure out ? 

4. Knowing the specific gravity of any given substance, how 
can you calculate the weight corresponding to any given meas- 
ure, or the measure corresponding to any given weight ? Give 
a general algebraic formula for the purpose, representing spe- 
cific gravity, weight, and volume, by Sp. Gr., W., and V. 

6. Determine the Sp. Gr. of absolute alcohol from the 
following data : — 

Grammes. 

Weight of bottle empty, 4.326 

" " fiUed with water at 4°, 19.654 

« « " alcohol at 0°, 16.741 

6. Determine the Sp. Gr. of lead shot from the following 
data : — 

Grammes. 

Weight of bottle empty, 4.326 

" " « filled with water at 4°, 19.654 

« " shot, 15.456 

" " bottle, shot, and water, 33.766 

7. Determine the Sp. Gr. of iron from the following data : — 

Grammes. 

Weight of iron in air, 3.92 

" " « water, 3.42 



WEIGHING AND MEASUKING. 41 

8. Determine the Sp. Gr. of copper from 

Grammes. 

Weight of copper in air, 10.000 

« " « water, 8.864 

9. Determine the Sp. Gr. of ash wood from 

Grammes. 

Weight of wood in air, 25.350 

" a copper sinker, 11.000 

" wood and sinker under water, 5.100 

10. How much bulk must a hollow vessel of copper fill, 
weighing one kilogramme, which will just float in water ? 

11. How much bulk must a hollow vessel of iron occupy, 
weighing ten kilogrammes, which sinks one half in water ? 

12. An alloy of gold and silver weighs ten kilogrammes 
in the air, and 9.735 kilogrammes in water. What are the 
proportions of gold and silver ? Sp. Gr. of gold = 19.2, of 
silver = 10.5. 

Solution. — In the French system the volume of a solid in cubic centimetres 

w 
equals its weight in grammes divided by its Sp. Gr., or V = g — q^. Since one 

cubic centimetre of water weighs one gramme, the volume of a solid in cubic 

centimetres is equal to the weight of water it displaces in grammes. Hence the 

w 
weight of water displaced = g — ^. Put, then, x = weight of gold in alloy, 

10 — X will equal weight of silver, y^ = weight of water displaced by gold, 

and -jjy^ = weight of water displaced by silver. Hence -^ + ~^ = 0.265. 

13. An alloy of copper and silver weighs 37 kilogrammes 
in the air, and loses 3.666 kilogrammes when weighed in water. 
What are the proportions of silver and copper ? 

22. Problems on Expansion of Liquids and Gases. 

1. If 34.562 cubic centimetres of mercury at 0° are heated 
to 100°, what increase of volume do they undergo, and what 
is the increased volume ? 

Solution. — The small fraction of its volume by which one c. c. of a liquid 

4* 



42 WEIGHING AND MEASURING. 

or gas increases ■when heated from 0° to 1°, is called the coefficient ofeocpandon 
of that liquid or gas. The coefficient of expansion of mercury, for example, 
= 0.00018, that is, one c. c. of mercury at 0° becomes 1.00018 c. c. at 1°. If 
Tve assume that the expansion is proportional to the temperature, then one 
c. c. at Qo becomes 1.00036 at 2°, 1.0009 at 5°, and 1.018 at 100°. Hence, 
84.562 c. c. of mercury would become, at 100°, (34.562 X 1.018) c. c. To make 
this solution general, let h = coefficient of expansion ; then (1 + ^) = increased 
volume of one c. c. when heated from 0^ to 1°, and (1 + < h) = increased 
volume of one c. c. when heated from 0° to <°, and V (1 + < ^) = increased 
volume of V c. c. when heated from 0° to t° ; representing by V the increased 
volume, we have 

Y'=Y {1 + toli), 

from which the increased volume of any liquid or gas may be calculated when 
the volume at 0°, the coefficient of expansion, and the temperature are knoAvn. 
It is not true, as we have assumed, that liquids expand twice as much for two 
degi'ees, three times as much for three degrees, &c., as they do for one; but, 
on the contrary, the rate of expansion slowly increases with the temperature. 
For example, one c. c. of mercury at 0° becomes 1.000179 at 1°, but one c. c. 
at 300o becomes 1.000194 at 331°. This difference of rate, however, is so 
small, that we can neglect it, except in the most refined experiments, more 
especially if we use not the coefficient observed at any particular temperature, 
but a mean coefficient obtained by observing the total amount of expansion 
between 0° and 100°, and then dividing the result by 100, by which we averrge 
the error. Again, experiments on the expansion of fluids are commonly con- 
ducted in glass vessels, which expand themselves by heat, and therefore cause 
the expansion to appear less than it is. If they expanded as much as the 
fluid, it is evident that the fluid would not appear to expand at all. They in 
fact expand much less than the fluids, but, nevertheless, sufficiently to make 
a material difference between the absolute expansion of a fluid, and its apparent 
expansion in glass vessels. The mean absolute coefficient of mercury between 
0° and 100° is 1.000181. The apparent expansion in glass between 0° and 
100° is 1.000156. In Table IV., at the end of the volume, will be found the mean 
coefficients of expansion in glass of some of the more important fluids. The 
rate of expansion of water varies so rapidly and so anomalously, that no use 
should be made of its coefficient except in experiments extending over 100°. 
When it is required to determine the amount of expansion between naiTower 
limits, use should be made of Table V., which gives the volume to which one 
cubic centimetre of water at 0° or at 4° increases, when heated to the tem- 
peratures at the side of the table. It also gives the Sp. Gr. of water at different 
temperatures, when either water at 0° or at 4° is taken as the unit. 

2. What will be the volume of 5.346 c. c. of water at 0°, 
when heated to 100° ? 

3. What will be the volume of 250 c. c. of oil of turpentine, 
at 0°, when heated to 50° ? 



WEIGHING AND MEASUKING. 43 

4. What will be the volume of 35 c. c. of water at 0°, when 
heated to 4°, to 25°, to 40°, and to 84° ? 

17. Problems on reducing Specific Gravities to the Standard 
Temperature. 

The specific gravities of both liquids and solids are supposed 
to be referred to water at 4°, its greatest density ; but in prac- 
tice we always use water at a much higher temperature, and it 
becomes therefore necessary to reduce the results to 4°, or in 
other words, to calculate what would have been the result had 
the temperature been 4° during the experiment. Hence an 
important class of problems like the following : — 

1. The Sp. Gr. of zinc was found to be 7.1582 when the 

temperature of the water was 15°. What would have been 

the Sp. Gr. at 4° ? 

Solution. — Sp. Gr. of water at 4° : Sp. Gr. at 15° = Sp. Gr. of zinc at 
15° : Sp. Gr. at 4°; or 1 : 0.9992647 = 7.1582 : x. 

2. The Sp. Gr. of antimony was found to be 6.681 when 
the temperature of the water was 15°. What would have 
been the Sp. Gr. at 4° ? Ans. 6.677. 

3. The Sp. Gr. of an alloy of zinc and antimony was found 
from the following data : — 

Grammes. 

Weight of the alloy, 4.4106 

" Sp. Gr. bottle, 9.0560 

" " " full of water at 4°, 19.0910 

« bottle, alloy, and water at 14°.6, 22.8035 

Ans. 6.375. 

4. Find the Sp. Gr. of metallic zinc from the following 
data : — 

Grammes. 

Weight of the zinc, 12.4145 

« bottle, 9.0560 

" « full of water at 18°, 19.0790 

" " zinc and water at 12°.4, 29.7663 

Ans. 7.153. 



44 WEIGHING AND MEASURING. 

24. Problems on reducing Centigrade Degrees to Fahrenheit, 
and the reverse. 

1. What do —40°, —20°, —15°, —9°, 0°, 4°, 10°, 13°, 15°, 
80°, 100°, 150° Cent, correspond to on the Fahrenheit scale ? 

Ride. — Double the number of degrees, subtract one tenth 
of the whole, and add 32 if the degrees are above 0°, or sub- 
tract 32 if they are below. 

2. What do —40°, —32°, —18°, —7°, 0°, 10°, 32°, 42°, 50°, 
70°, 90°, 100°, 150°, 212°, 300°, 450° Fahr. correspond to on 
the Centigrade scale ? 

Rule. — Subtract 32 if above 0°, or add 32 if below, add 
one ninth to the result, and divide by two. 

27. Problems on Expansion of Solids. 

1. What will be the length of a rod of iron 424.56 metres 
long at 0°, when heated to 20° ? 

Solution. — The small fraction of its length by which a rod of iron, or of any 
other solid, one metre long, expands when heated from 0° to 1°, is called the 
coefficient of linear expansion of the solid. A bar of iron one metre long 
at 0° becomes 1.0000122 at 1°, and the small fraction 0.0000122 is the 
coefficient of linear expansion of iron. Eepresenting this coefficient by h, 
Ave have for the new length of the rod, at 1°, V = 1 (1 + X"), and at io^ 
V = l{\-\-t'k). We may assume, in the case of solids, that the expansion is 
proportional to the temperature, especially if we deduce our coefficient from 
experiments made between 0° and 100°, as described under examples on the 
expansion of fluids. Substituting for ?, ^, and t the values given in the problem, 
we have 11 = 424.56 (1 + 20 x 0.0000122). The coefficients of linear expan- 
sion for a number of solids are given in Table IV. at the end of the volume. 

2. What will be the length of a rod of copper 2.365 metres 
long, at 0°, when heated to 100° ? 

3. What will be the length of a rod of silver 0.760 metres 
long, at 12°, when heated to 20° ? 

Solution. — Denoting by I the unknown length of the rod at 0°, hy I' the 
known length at i°, and I" the required length at t'o, we have as above, 

l'=l{l + tk), and l"^Hl + fk). 



WEIGHING AND MEASURING. 45 

By combining these, we obtain 

^" = ^ (rrfi) = ^' [1 + ^ «' - + &c.]. 

All the terms of the quotient may be neglected after the first, because they 
contain powers of the already very small fraction Ic. Substituting the values 
in the above equation, we get 

l' =z 0.760 [1 + 0.000019 (20 — 12)]. 

4. One of the large iron tubes of the Britannia Bridge over 
the Menai Straits is 143.253 metres long. What increase of 
length does it undergo between 8° and 20° Centigrade ? 

5. What is the increased capacity of a globe of glass which 
holds exactly one litre, at 0°, when heated to 250° ? 

Solution. — If we consider for a moment the glass globe as forming the out- 
side shell of a solid globe of glass, it is evident that the increased capacity of 
the globe will be equal to the increased volume of this solid globe, which we 
have supposed for a moment to fill the interior. The problem, therefore, 
resolves itself into calculating the amount of cubic expansion of a solid glass 
globe having a volume equal to one litre, or 1000 c. c. The coefficient of linear 
expansion of glass is given in the table as 0.00001. The coefficient of cubic 
expansion is always three times * as great as the linear expansion ; in the 
case of glass, therefore, it is 0.00003. By this is meant that, if a rod of glass 
one metre long at 0° becomes 1.00001 long at 1°, then a cube of glass of one 
cubic centimetre at 0° becomes 1.00003 c. c. at 1°. Using, then, the equa- 
tion 

we obtain by substitution 

V = 1000 (1 + 250 X 0.00003) = 1007.5 c. c. 

6. What is the increased capacity of a globe of glass which 
holds exactly 495 c. c. at 15°, when heated to 300° ? 



* Each edge of a cube of glass one centimetre long at 0°, would become 
(1 + h) c. m. long at 1°. The increased volume of the cube would be equal to 
(1 -)- Z;)3 = 1 + 3 ^- 4- 3 X-2 + 7j3; but as Ic is an exceedingly small fraction, 
Z;2 and Z;3 may be neglected in comparison, so that a cube of glass of one c. c, 
at 0° becomes {1 + Zh)^i 1°, which proves that the amount of cubic expan- 
sion is three times as great as the linear. 



NON-METALLIC ELEMENTS, or 
METALLOIDS. 



FIRST GEOUP : ORGANOGENS. 

Oxygen (0). 



56. Hg b = ^ + ©. 

1. How much oxygen can be obtained by heating 108 
grammes, 250 grammes, 25 grammes, or 5 grammes, of 
red oxide of mercury ? 

2. How many cubic centimetres of oxygen can be 
obtained from the amounts of oxide of mercury given 
in the last example ? 

3. How much oxide of mercury would be required to 
yield one litre of oxygen by the process of 1 ? 

4. How much mercury would remain after the experi- 
ment of last example ? 

59. KO, CIO5 = KCl+6©. 

1. How much oxygen can be obtained from 1 kilo- 
gi-amme, 50 grammes, or 5 grammes, of chlorate of po- 
tassa ? 

2. How much chloride of potassium would remain after 
the oxygen in the last examples had been driven off? 



NON-METALLIC ELEMENTS, OR METALLOIDS. 47 

3. How much cMorate of potassa would be required to 
make one litre of oxygen ? 

63. C + 2 O = C O2. 

1. How much carbonic acid gas will be formed by burn- 
ing 5 grammes of carbon ? 

2. How many cubic centimetres of carbonic acid will 
be formed by burning 5 grammes of carbon ? 

3. How much oxygen will be consumed in the last two 
examples ? 

4. How many cubic centimetres of carbonic acid will 
be formed from one litre of oxygen, by burning in it 
carbon ? 

64. S + 2 O = S O2. 

1. How much sulphurous acid gas wiU be formed by 
burning 10 grammes of sulphur ? 

2. How much oxygen will be consumed in the last 
example ? 

3. How many cubic centimetres of sulphurous acid wiU 
be formed by burning sulphur in one litre of oxygen ? 

4. Assuming that one litre of oxygen yields exactly one 
litre of sulphurous acid, what is the Sp. Gr. of S Og gas ? 

65. P + 50 = P05. 

1. How much phosphoric acid can be formed from 48 
grammes of phosphorus ? 

2. How much phosphorus will exactly consume one 
litre of oxygen ? 

3. How much phosphorus is required in order to make 
250 grammes of phosphoric acid ? 



48 NON-METALLIC ELEMENTS, OR METALLOIDS. 

67. NaH-© = NaO. 

1. How much oxide of sodium can be made from 
28.75 grammes of sodium ? 

2. How much oxide of sodium can be made with one 
litre of oxygen ? and how mucli sodium will be consumed 
in the experiment ? 

68. 3 Fe + 4 O = Fe O, Fea O3. 

71. 2NaO-^POs-^Aq= HO, 2NaO,PO,-{- Aq. 

The ciystallized salt = [HO, 2NaO] PO5 . 24HO. 

72. Ca0^aO^^Aq= CaO, CO2 + Aq. 

73. CaO-{- SOi-\-Aq= Ca 0,S0^-\- Aq 

1. How much lime would be required to neutralize 20 
grammes of sulphurous acid ? 

2. How much lime would be required to neutralize 
the sulphurous acid obtained by burning 5 grammes of 
sulphur ? 

3. How much lime would be required to neutrahze one 
litre of sulphurous acid gas ? 

79. 3 Mn Oi = Mn O, Mng O3 + 2 O. 

1. How much oxygen can be obtained from one kilo- 
gramme of hyperoxide of manganese by heating ? 

2. How much of the oxide must be used in order to 
obtain 30 grammes of oxygen ? 

IVInOa + HO, ^S C>3 = MnO, S O3 4- HO + O. 

1. How much oxygen can be obtained from one kilogr. 
of hyperoxide of manganese by the last process ? and how 
much sulphuric acid wiU be required to decompose it ? 

2. "What per cent of the whole amount of oxygen in 
the mineral is obtained by the two processes ? and by how 
much does the second exceed the first ? 



NON-METALLIC ELEMENTS, OR METALLOIDS. 49 

Hydrogen (H). 

81. Na + ^g = Na 0, HO^Aq-\- H. 

1. How much hydrogen will be set free by 5.25 
grammes of sodium? How many cubic centimetres will 
be set free? 

82. 3 Fe + 4 M © == Fe O, Fcj O3 + 4 H. 

1. How much hydrogen would be obtained in the last 
experiment by the decomposition of 39 grammes of 
water ? 

2. By how much would the weight of the tube increase 
in last example ? 

83. Zn + HO, S 0^^ Aq = Zn 0, S 0^-^ Aq -\-ll. 

1. How much sulphuric acid and how much zinc must 
be used in order to make 2 grammes of hydrogen ? How 
much in order to make one litre ? 

2. How much hydrogen can be obtained with one kilo- 
gramme of zinc ? 

3. How much with one kilogramme of sulphuric acid ? 

Decomposed by Galvanism. 

55. iro = H + ©. 

1. How many cubic centimetres of hydrogen would be 
obtained by the decomposition of one cubic centimetre 
(one gramme) of water ? How many cubic centimetres 
of oxygen ? How many of mixed gases ? 
85. A solid immersed in a liquid or a gas is buoyed up 
by a force equal to the weight of the liquid or gas which 
it displaces. The excess of the buoyancy over its own 
weight is called its ascensional force. 

1. What would be the ascensional force of a small 
balloon filled with one htre of hydrogen gas, when the 
balloon itself weighs five centigrammes ? 
5 



50 NON-METALLIC ELEMENTS, OR METALLOIDS. 

2. What would be the ascensional force of a spherical 
balloon seven metres in diameter, two thirds filled with 
hydrogen, when the balloon and attachments weigh twenty 
kilogrammes ? 

87. n-i-o = iro. 

1. How much water would be formed by burning one 
thousand litres of hydrogen? and how much oxygen 
would be consumed in the process ? 

2. How much vapor of water would be formed in Ex- 
ample 1 ? 

93. Problems on the Barometer. 

1. When the surface of the column of mercury in a 
barometer stands at 76 centimetres above the mercury in 
the basin, with what weight is the atmosphere pressing on 
every square centimetre of surface ? Sp. Gr. of mercury 
= 13.596. 

2. To what difference of pressure does a difference of 
one centimetre in the barometric column correspond ? 

3. When the water barometer stands at ten metres, 
what is the pressure of the air if the temperature is 4° ? 

4. How high would an alcohol barometer, and how 
high a sulphuric-acid barometer, stand under the same 
circumstances, disregarding in each case the tension of 
the vapor ? Sp. Gr. of alcohol = 0.8095 ; Sp. Gr. of 
sulphuric acid = 1.85. 

94. Prohlems on the Compression and Expansion of Gases. 

Mariotte's Law. — It is an estabhshed principle of sci- 
ence that I'he volume of a given weight of gas is inversely 
as the pressure to which it is exposed ; that is, the greater 
the pressure, the smaller is the volume ; and the less the 



NON-METALLIC ELEMENTS, OR METALLOIDS. 51 

pressure, the larger is the volume. This may be illustrated 
by an India-rubber bag holding one litre of air, or of any 
other gas. This is exposed to a pressure, under the ordi- 
nary conditions of the atmosphere, of a little over one 
kilogramme on every square centimetre of surface. If 
this pressure is doubled, the volume of the bag will be 
reduced to one half; if trebled, to one third, &c. On 
the other hand, if the pressure is reduced to one half, 
the volume will double ; if to one third, the volume will 
treble, &c.* The principle is expressed in mathematical 
language by the proportion 

- H' : H=: V : V; (1.) 

where H and H' are the heights of the barometer which 
measure the pressure to which the gas is exposed under 
the two conditions of volume V and V. 

Since the density of a given weight of gas is inversely 
as the volume, or D' : D = V : V, it follows that 

H' : H = D' : D, (2.) 

or the density of a gas is proportional to the pressure to 
which it is exposed. Moreover, since the weight of a 
given volume of gas is proportional to its density, and its 
density, as just proved, proportional to the pressure, it 
follows that The weight of a given volume of gas is directly 
as the pressure to which it is exposed, or 

H' : H = W : W. (3.) 

These three proportions are very important, and will be 
constantly referred to in the following pages. The student 
must be careful to notice that in (1) the weight of gas is 
supposed to be constant and the volume to vary, and in 
(g) the volume is supposed to be constant and the weight 
to vary. 

* We suppose the bag to have no elasticity. 



52 NON-METALLIC ELEMENTS, OR METALLOIDS. 

The variations in the pressure of the atmosphere, amount- 
ing at times to one tenth of the whole, necessarily cause 
equally great changes in the volume of gases which are 
the objects of chemical experiment. Hence, in order to 
compare together different volumes of gas, it is essential 
that they should have been measured when exposed to 
the same pressure. A standard pressure has therefore 
been agreed upon, that measured by 76 centimetres of 
3iiercury, to which the volume of gases measured under 
any other pressure should be reduced. Hence a number 
of problems like the following : — 

1. A volume of hydrogen gas was measured and found 
to be equal to 250 c. c. The height of the barometer, 
observed at the same time, was 74.2 centim. What 
would have been the volume if observed when the barom- 
eter stood at 76 centim. ? 

Solution. — Proportion (1) gives, by substituting the data of the prob- 
lem, 74.2 : 76 = 250 : V = Ans. 

2. A volume of nitrogen gas measured 756 c. c. when 
the barometer stood at 77.4 centim. What would it have 
measured if the barometer had stood at 76 centim. ? 

3. A volume of air standing in a bell-glass over a mer- 
cury pneumatic trough measured 568 c. c. The barome- 
ter at the time stood at 75.4 centim., and the surface of the 
mercury in the bell was found, by measurement, to be 6.5 
centim. above the surface of the mercury in the trough. 
What would have been the volume had the air been ex- 
posed to the pressure of 76 centim. ? 

Solution. — It can easily be seen, that the pressure of the air on the 
surface of the mercury in the pneumatic trough, measured by the 
height of the barometer at the time (75.4 centim.), was balanced first by 
the column of mercury in the bell, and secondly by the tension * of the 
confined air. Hence, the pressure * to which the air was exposed was 

* The tension of a gas is the force with which it tends to expand, and, when 
the gas is at rest, must evidently be exactly equal to the pressure to which it 
is exposed. 



NON-METALLIC ELEMENTS, OR METALLOIDS. 53 

equal to the height of the barometer less the height of the mercury 
in the bell, or 75.4 — 6.5 = 68.9 centim. We have then the proportion 
68.9 : 76 = 568 : V = Ans. 

4. A volume of air standing in a tall bell-glass over a 
mercury pneumatic trough measured 78 c. c. The barom- 
eter at the time stood at 74.6 centim., and the mercury in 
the bell at 57.4 centim. above the mercury in the trough. 
"What would have been the volume had the pressure been 
76 centim. ? 

5. What would be the answers to the last two problems, 
had the pneumatic trough been filled with water instead 
of mercury ? 

97. Problems on Expansion of Gases hy Heat. 

1. What will be the volume of 250 cubic centimetres 
of air at 0° when heated to 300° ? 

Solution. — It has been found by Eegnault and others, that the per- 
manent gases expand so nearly equally for the same increase of tempera- 
ture, that the differences may be entirely disregarded except in the most 
refined investigations; and it has also been found, that their rate of 
expansion does not materially vary from the lowest to the highest tem- 
peratures at which experiments have been made. The coefficient of 
expansion for air, as determined by Eegnault, is equal to 0.00366, and 
we can therefore calculate the volume, V, of a gas at any temperature, 
from its volume, V, at 0°, by means of the equation, already explained, 

V' = V(l + iX 0.00366); (1.) 

or if we know the volume Y' for a tempei-atiire t, we can calculate the 
volume V" for another temperature i\ by means of the equation 

V" = V' (! + («' — <) 0.00366). (2.) 

By transposing, we can obtain from equation (1) 

l + tX 0.00366' ^ ' 

As the volume of a gas varies very considerably with the temperature, 
it is important, in comparing together different measurements, that we 
should adopt a standard temperature, as we have adopted a standard 
pressure. The temperature which has been agreed upon is 0° ; but as it 
would be inconvenient, and often impossible, to make our measurements 
at this temperature, it becomes necessary to calculate, by means of equa- 
5* 



54 NON-METALLIC ELEMENTS, OR METALLOIDS. 



tion (3), from a volume Y' measured at t°, what would be the volume at 
0°. This is called technically reducing the volume to 0°. There can be 
obtained also from equations (1) and (2) the equations 

_ V'' — V , _ V" — V 

' ~ V X 0.00366 ' ^^-^ ^^^ * ~ * "^ V X 0.00366 ' ^^'^ 

by means of which we can calculate the change of temperature when 
we know the change of volume. Kepresenting the coefficient of expan- 
sion in the above formulte by Jc, we can obtain, by transposing and re- 
ducing the equation, 

V'-V , , , V" — V 

'' = -lir' (60 and ^ = v' (£/-«)• CO 

From these we can calculate the coefficient of expansion when we know 
the volume of a gas at two diflfereat temperatures. 

2. A volume of gas measured 560 c. c. at 15°. What 
would it measure at 95° ? 

3. A glass globe holding 450 c. c. of air at 0° was 
heated to 300°. At this temperature the neck was her- 
metically sealed, and the globe cooled again to 0°. The 
neck was then opened under mercury, and the air remain- 
ing in the globe passed up into a graduated jar, and meas- 
ured. How much was it found to measure ? 

Solution. — By substituting the values for V and t in the equation 

V = V (1 + < X 0.00003), we obtain the increased capacity of the globe, 
and of course the number of cubic centimetres of expanded air which it 
contains at 300^. It is then only necessary to substitute this value for 

V and 300 for t, in equation (3), in order to find what will be the volume 
of this expanded air when cooled again to 0°. 

4. "What is the weight of air contained in an open glass 
globe of 250 c. c. capacity, at the temperature of 20°, and 
when the barometer stands at 74 centimetres ? 

Solution. — In order to make the solution general, we will represent the 
capacity of the globe, the temperature, and the height of the barometer, 
by V, t, and H, respectively. One cubic centimetre of air at 0°, and 
when the barometer stands at 76 centimetres, weighs 0.00129 grammes. 
To find what one cubic centimetre would weigh when the barometer 
stands at H centimetres, we make use of proportion (3), page 51: 

H : H' = W : W; or 76 : H = 0.00129 : W; 

whence W = 0.00129 . ^, the weight of one cubic centimetre at 0^, and 
under a pressure of H centimetres. To find what one cubic centimetre 



NON-METALLIC ELEMENTS, OR METALLOIDS. 



55 



would weigh at P, it must be remembered that one cubic centimetre at 0= 
becomes (1 ■+• t 0.00366) c. c. at t^; therefore, at P and at H centimetres 
of the barometer, (1 + i 0.00366) c. c. weigh 0.00129 . ^ grammes. By 
equating these two terms we obtain (1 + 1 0.00366) = 0.00129 . ^, whence 
1 = 0.00129 • ]-^ x7o"oo366 ' 76' *^® weight of one cubic centimetre at P and 
under a pressure of H centimetres. The weight of V cubic centimetres, 
w, is evidently 



zi) = 0.00129 V . 



H 



1 + ( 0.00366 76 * 



(8.) 



5. Wliat is the weight of air contained in an open glass 
globe of 560 cubic centimetres' capacity at 0°, at the 
temperature of 300°, and under a pressure of 77 centi- 
metres ? 

Solution. — In the soli;tion of the last example we neglected the change 
of capacity of the glass globe due to the change of temperature. This 
causes no sensible error when the change of temperature is small, but 
when, as in the present problem, the change of temperature is quite 
large, the change of capacity of the globe must be considered. If the 
capacity is V c. c. at 0^, it becomes atPY{l + t 0.00003). Introducing 
this value for V into the equations of the last section, we obtain 



w = 0.00129 V (1 + < 0.00003) 



1 + t 0.0U366 76 ■ 



(9.) 



99. Problems on Specific Gravity of Vapors. 

General Solution. — The specific gravity of a vapor is its weight com- 
pared with the weight of the same volume of air under the same con- 
ditions of temperature and pres- 



sure. To find, then, the specific 
gravity of a vapor, we must as- 
certain the weight of a known 
volume, V, at a known tempera- 
ture, t, and under a known pres- 
sure, H, and divide this by the 
weight of the same volume of air 
at the same temperature, and 
under the same pressure. The 
method may best be explained by 
an example. Suppose, then, that 
we wish to ascertain the specific 
gravity of alcohol vapor. We take 
a light glass globe having a ca- 
pacity of from 400 to 500 c. c, and 



Fig. 6. 




56 NON-METALLIC ELEMENTS, OR METALLOIDS. 

draw the neck out in the flame of a blast lamp, so as to leave only a fine 
opening, as shown in the figure at a. The first step is now to ascertain 
the weight of the glass globe when completely exhausted of air. As 
this cannot readily be done directly, we weigh the globe full of air, and 
then subtract the weight of the air, ascertained by calculation from the 
capacity of the globe, and from the temperature and pressure of the air, 
by means of equation (8). Call the weight of the globe and air W, and 
the weight of the air w, then W — w is the weight of the globe exhausted 
of air. The second step is to ascertain the weight of the globe 
filled with alcohol vapor at a known temperature, and under a known 
pressure. For this purpose we introduce into the globe a few grammes 
of pure alcohol, and mount it on the support represented in Fig. 6. By 
loosening the screw, r, we next sink the balloon beneath the oil contained 
in the iron vessel, V, and secure it in this position. We noAV slowly raise 
the temperature of the oil to between 300° and 400°, which we observe by 
means of the thermometer, B. The alcohol changes to vapor and drives 
out the air, which, with the excess of vapor, escapes at a. When the 
bath has attained the requisite temperatui'e, we close the opening a, by 
suddenly melting the end of the tube at a with a mouth blow-pipe, and 
as nearly as possible at the same moment observe the temperature of the 
bath and the height of the barometer. We have now the globe filled with 
alcohol vapor at a known temperature, and under a known pressure. 
Since it is hermetically sealed, its weight cannot change, and we 
can therefore allow it to cool, clean it, and weigh it at our leisure. 
This will give us the weight of the globe filled with alcohol vapor 
at a known temperature, t', and under a known pressure, H'. Call 
this weight W'. The weight of the vapor is W' — W + w. The 
third step is to ascertain the weight of the same volume of air at the 
same temperature and under the same pressure. This can easily be 
found by calculation from equation (9). The last step is to find the 
capacity of the globe, which, although we have supposed it known, is 
not actually ascertained experimentally until the end of the process. 
For this purpose we break off the tip of the tube (a), under mercury, 
which, if the experiment has been carefully conducted, rushes in and 
fills the globe completely. We then empty this mercury into a carefully 
graduated glass cylinder, and read oS" the volume. We find then the 
specific gravity by dividing the weight of the vapor by the weight of 
the air. The formula for the calculation are then 

Weight of the globe and air, W. 

" " air, ,„ = 0.00129 V.pp-^^. 5. 

" " globe exhausted of air, W — w. 

" " " filled with vapor '\ 

at a temperatm-e t' and under a > W. 

pressure H', ) 



NON-METALLIC ELEMENTS, OR METALLOIDS. 



57 



Weight of the vapor, W — W + to. 

" " air a.t t' ) 1 H 

ail HI I ( =,0.00129 V(l + «' 0.00003)- i + ,. o.uosss ' 76 ' 
jsure H', ) ^ 

W — W + w 

Sp. Gr. = . 



and under a pressure 



0.00129 V (l + C 0.00003). 



1 + t 0.00366 75 



1. Ascertain the Sp. Gr. of alcohol vapor from the fol- 
lowing data : — 



"Weight of glass globe, W 


50.8039 grammes. 


Height of barometer, H 


74.754 centim. 


Temperature, t 


18° 


Weight of globe and vapor, W 


50.8245 grammes. 


Height of barometer, H' 


74.764 centim. 


Temperature, i' 


167° 


Volume, V 


351.5 cubic centim. 




Ans. 1.5795. 


2. Ascertain the Sp. Gr. of 


camphor vapor from the 


following data : — 




Weight of glass globe, W 


50.1342 grammes. 


Height of barometer, H 


74.2 centim. 


Temperature, t 


13°.5 


Weight of globe and vapor, W 


50.8422 grammes. 


Height of barometer, H' 


74.2 centim. 


Temperature, t 


244° 


Volume, V 


295 cubic centim. 




Ans. 5.298. 



Carbon (C). 

109. C + 20 = C02. 

2 Hg 4- C = 2 ^r/ -}- C ©2- 

1. HoAV many grammes and how many cubic centi- 
metres of carbonic acid gas are formed by burning 10 
grammes of charcoal ? 



58 ■ NON-METALLIC ELEMENTS, OR METALLOIDS. 

2. Ho"W many grammes and hoTV many cubic centi- 
metres of oxygen are consumed in the process ? 

3. Assuming that the volume of carbonic acid gas gen- 
erated during combustion is exactly equal to the volume of 
oxygen gas consumed, what is the Sp. Gr. of carbonic 
acid gas ? 

4. How much oxide of mercury is required to bum up 
5.672 grammes of charcoal ? 

110. C + = €0. C + C02=2CO. 

1. How many grammes and how many cubic centime- 
tres of oxide of carbon gas are formed by burning 10 
grammes of charcoal ? 

2. How many grammes and how many cubic centime- 
tres of oxygen are consumed in the process ? 

8. Ten cubic centimetres of oxygen yield how many 
cubic centimetres of carbonic acid gas, and how many of 
oxide of carbon gas ? What expansion does oxygen un- 
dergo in combining with carbon to form oxide of carbon ? 

Spermaceti. 

115. C64 H64 O4 + 188 O = 64 C O2 + 64 HO. 

1. How many grammes of carbonic acid, and how many 
grammes of water, are formed by burning 10 grammes of 
spermaceti ? 

2. The carbonic acid and water given off by a burning 
spermaceti candle were carefully collected and weighed. 
The water weighed 0.564 grammes, the carbonic acid 
weighed 1.3786. How much of the candle was burned? 



NON-METALLIC ELEMENTS, OR METALLOIDS. 59 

SECOND GROUP OF METALLOIDS. 

Sulphur (S). 

132. Fe S + HO, 80^-\-Aq^Fe0,S0^^Aq•\- H S. 

1. How much sulphide of hydrogen can be made from 
15 grammes of sulphide of iron ? How many cubic centi- 
metres ? 

2. How much sulphide of iron, and how much sulphuric 
acid, is required to generate sufficient gas to saturate one 
litre of water ? 

HS-^Aq-\-i^=^-\-Aq. 
HS + 30= eO + SOij. 

Black. 

133. a. Pb + iT^-f ^^ = Pb S + ^^ + H. 

Yellow. Black. 

l.YhO ^ H S -\- Aq = Y'o^ -\- Aq. 

Black. 

c.PlO, {_G^H;\ O3 4- H8 -\- Aq = PbS -f 
HO,lG^H,-\0,JrM- 

Black. 

d. Fe 0,S0^-\- GaO,HO-\- HS -{-Aq^YQ^ + 
Ga 0,S0^-\- Aq. 

Black. 

Gu 0, [O; H,-] 0, -{. ITS + Aq = Cu S + 
irO,[G,ff,-] 0, + Aq. 

Orang'e# 

Sb Gl3-}-3JIS-\-Aq = ^hS3-{-3irGl-{- Aq. 

Yellow. 

As Gls -i-SIIS-^ Aq = As S3 -\- dHGl-\-Aq. 

White. 

ZnGl-{-GaSA Aq =ZnS-\- Ga Gl-\- Aq. 



60 NON-METALLIC ELEMENTS, OR METALLOIDS. 

Phosphorus (P). 

140. P + 3 O = P O3 (by slow combustion). 
P-f-5O==P05(by rapid combustion). 

144. Burnt bones consist chiefly of 3 Ca O, cP O5. 

3 Ca O, cP O5 + 2 {HO, S0,)-\-aq = 2 (CaO, S O3) 
+ [2 HO, Ca 01 eP O5 + aq. 

Heated to a red heat. 

\^H0, Ca 0], ^PO, ^ Aq -\-xC = CaO, PO5 

+ a;C + Aq + 2CO + 2H. 

Heated intensely. 

3 (CaO, ,P Os) + a; C = 3 CaO, aPOs + a: C + 2 P 
+ 10 C O. ^ 

1. How much phosphorus can be manufactured from 
20 kilogrammes of burnt bones, of which four fifths are 
phosphate of lime ? 

145. 4P + 3(CaO, HO) = 3(CaO, P0) + PH3. 

Besides the above reaction, there take place simulta- 
neously the two following reactions, in the experiment 
described in the text-book. 

3 P 4- 2 (Ca 0, H 0) = 2 (Ca 0, P 0) + P H^. 

P -f Ca O, H = Ca 0, P + M. 



THIRD GROUP OF METALLOIDS. 

Chlorine (CI). 

150. Mn 0.,-^2HGl-\-aq = Mn CI -{- aq -{- CI. 

1. HoAv much chlorine gas can be obtained from 2.467 



NON-METALLIC ELEMENTS, OR METALLOIDS. 61 

grammes of clilorohydric acid gas ? How many cubic 
centimetres ? 

2. How much chlorine can be obtained from an unde- 
termined amount of muriatic acid by means of 4.567 
grammes of hyperoxide of manganese ? How many cubic 
centimetres ? 

3. The hyperoxide of manganese of commerce is more 
or less adulterated. What per cent of Mn O2 does an 
article contain, of which 10 grammes, when heated with 
strong muriatic acid, evolve 4.0135 grammes of chlorine ? 

4. How much chlorine can be obtained from 25 cubic 
centimetres of muriatic acid* of Sp. Gr. = 1.16 ? How 
many cubic centimetres ? 

5. In order to prepare one litre of chlorine gas how much 
hyperoxide of manganese, and how much muriatic acid, 
must be used ? Calculate the amounts for pure Mn O2 and 
H CI gas, and also when the oxide used contains only 
70 per cent of pure MnOa, and when the liquid acid 
used has a Sp. Gr. = 1.15. 

151. Mn O2 + 2 Na CI + 2 (^ 0, aS' O3) -{- aq = Mn a + 
2 (JVa 0, S Os) -{- ag -{- Ch 

"We might use one half as much common salt, but then 
we should find sulphate of manganese instead of chloride 
of manganese in solution. Thus, 

Mn O2 + Na CI + 2 {HO, S 0^) -{-aq=-.Mn 0, SO, 
Jf-J^aO,SOs-\-aq^ CI. 

1. How much chlorine gas can be obtained by the last 
process from 34 kilogrammes of salt ? 

* See Table VI., which gives the per cent of H CI in the fluid acid of dif- 
ferent specific gravities. 

6 



62 NON-METALLIC ELEMENTS, OR METALLOIDS. 

2. How many cubic centimetres of chlorine can be ob- 
tained from one cubic centimetre of rock salt ? Sp. Gr. 
of salt = 2.15. 

152. /. 2 {Fe 0, S 0^) -{- H 0, S 0^ -\- Gl -\- Aq = Fe^ 0^, 
3 S Os + II CI -\- Aq. 

6 (Fe 0,8 0^) -\- 3 CI -\-Aq = 2 {Fe.^ O3, 3 S O3) 
_|_ Fe^ Ck + Aq. 

g. An -^ 3 CI -\- Aq =^ An CI, + Aq. 



ACIDS 



FIRST GROUP : OXYGEN ACIDS. 

Nitrogen and Oxygen. 

Nitric Acid {HO, NO,). 
159. KO, N Os + 2 {HO, 8 0^) = KO, SO,. HO, S O3 

NaO, NO5 + 2 {HO, SOs) = NaO, SO,. HO, S 0, 

+ H0,J¥05. 

1. How much nitric acid can be made from 250 kilo- 
grammes of potash nitre, and how much sulphuric acid 
must be used in the process ? 

2. How much more nitric acid will the same weight of 
soda nitre yield ? 

3. How much nitric acid, containing 40 per cent, of 
NO5, can be made from 1700 kilogrammes of potash 
nitre ? 

4. How much soda nitre, and how much sulphuric acid, 
and how much water, must be used to make 450 kilo- 
grammes of nitric acid, which shall contain 60 per cent. 
of pure acid ? 



64 ACIDS. 

Ammonia, 

160. c. INH,-] 0, HO + HO, NO, -^Aq^ iNH,-\ 0, N 0, 

+ Aq. 

d.VhO-{-HO,NO,-\-Aq = PhO,NOs-\-Aq. 

1. How much nitric acid, of Sp. Gr. 1.14, is required 
to dissolve 20 kilogrammes of oxide of lead ? 

2. How much nitric acid, of Sp. Gr. 1.14, and how 
much oxide of lead, must be used to make 10 kilogrammes 
of nitrate of lead ? 

c. 3 Pb + 4 {HO, N O5) -\- aq =^ Z (Pb 0, NO5) + aq 

+ 1V02. 

3 Cu + 4 {HO, NO,)^aq= 3 {Cu 0, NO,) + aq 

1. How much nitric acid, of Sp. Gr. 1.22, is required 
to dissolve 450 grammes of lead ? How much nitrate of 
lead would be formed ? 

2. How much nitric acid, of Sp. Gr. 1.362, is required 
to dissolve 450 grammes of copper ? 

/. 3 P + 5 (^ (9, NO,) -{-Aq=3{3HO,,P 0,) + Aq 
+ 5 ]^ O2. 
S + HO, NO, -\-Aq = HO,SOs + Aq + JVO^. 

Nitric Oxide (X O). 

6FeCl-\-KO,N05-{- 4.H01 -{- Aq = S Fe^ CI3 + 

KCl + aq-\-NO!i. 

Colorless. Red. 

1. How much IV O2 can be obtained by dissolving 10 
grammes of copper in nitric acid ? How many cubic 
centimetres ? 



ACIDS. 65 

2. How much IV O2 can be obtained from 10 grammes 
of iron by the last reaction but one ? 

3. "What volume of oxygen must be mixed with one 
litre of TV O2 in order to change it into ]V O4 ? 

Nitrous Oxide (IV O). 

When heated. 

163. [NHi]0, N05=4HO + 2]VO. 

Sp.Gr. of 1.14. 

4.Vh^b{H0, N0,)+Aq = 4: (Pb 0, NOs) + Aq 

+ JVO. 

Mi 0, S 0^ -\- Aq -{- -X 0^ = Na 0, S 0^ -{- Aq -\- 
IV©. 

1. Ten grammes of nitrate of ammonia yield how many 
grammes and how many cubic centimetres of protoxide of 
nitrogen ? 

2. How much nitrate of ammonia must be used in order 
to make one litre of the gas ? 

3. One litre of TV O2 yields how many cubic centime- 
tres of IV O by the third reaction of this section ? 

4. One litre of IVO gives, when decomposed, what 
volume of nitrogen ? 

Carbon and Oxygen. 
Carbonic Acid (C 02)' 

164. CaO, CO2 4- HO, NO, -\- Aq = Ca 0, NO^ + Aq 

+ CO2. 

Ca 0, C O2 + ^0, S 0^-\- Aq= C^O, S O3 + Aq 

+ C O2. 

6* 



•66 ACIDS. 

Ca 0, NOs + HO, SO3 + aq = C?.0, ^0^-{- 
HO,NOs-\-aq. 

1. How much sulphuric acid and how much nitric 
acid* must be used to drive out all the carbonic acid from 
25.462 grammes of chalk ? How many grammes and how 
many cubic centimetres of gas would be obtained ? 

2. The specific gravity of Carrara marble is 2.716. 
How many cubic centimetres of carbonic acid gas does 
one cubic centimetre of the marble contain in a condensed 
state ? 

167. 1. Animals remove oxygen from the air, and return the 
whole as carbonic acid. Plants remove carbonic acid, 
and, having decomposed it, return the oxygen it contained. 
How does the volume of the oxygen in either case com- 
pare with that of the carbonic acid ? 

Sulphur and Oxygen. 
Sulphuric Acid (S O3). 

169. S02 4-Ot= SO3. 

1. How much anhydrous sulphuric acid is formed by 
the oxidation of 10 grammes of sulphurous acid? and 
how much oxygen is required in the process ? 

2. How much anhydrous sulphuric acid is formed by 
the oxidation of one litre of sulphurous acid gas ? and 
how many cubic centimetres of oxygen must be mixed 
with it in the experiment ? 

a. 2 S O3 + H O = iT 0, 2 ^ ^3 (the Nordhausen Acid). 

* When the strength of the nitric acid is not stated, monohydrated acid 
(H 0, N O5) is always intended. 

t The two gases are mixed together and led over heated platinum sponge in 
a glass tube. 



ACIDS. 67 

S O3 + H O = ^ 0, aS' 6>3 (the common Acid). 

170. Fe O, S O3 . 6 H is the symbol of ciystallized green 

vitriol. 

When healed. 

2(FeO, SO3 . 6HO) = Fe203, S O3 + 12IIO 

+ S©2. 
By further heating", 

Fe2 03, S03=Fe2 03 + §©3. 

By conducting the anhydrous acid fumes into HO, S O3 
we get IIO,2S03. 

1. How much anhydrous acid, and how much Nord- 
hausen, can be made by the above process from 20 kilo- 
grammes of green vitriol ? If the Nordhausen acid has 
the specific gravity of 1.9, how many litres can be obtained 
from 20 kilogrammes of green vitriol ? 

171. S02 + M.0,N0s + Afl = HO, SOs + Aq-^NOi. 

1. How much HO, S O3 will be formed from one 
gramme of S O2 ? How much from one litre ? 

While. 

Ba Ol-{- HO, S03-\-Aq = BiiO,SO3-^HCl-\- Aq. 

BaO,NOs + HO, S 0^ -\- Aq = BaO, SO3 + 
HGl-\-Aq. 

1. Why must sulphuric acid or a soluble sulphate pro- 
duce a precipitate when added, in solution, to the solution 
of any salt of baryta ? 

2. The precipitate produced by adding an excess of 
Ba CI to a solution of HO, S O3 was collected, and 
weighed 4.567 grammes. How much sulphuric acid* 
was present in solution ? 

* When the name sulphuric acid is used, H 0, S O3 is always meant, unless 
otherwise specified. 



69 ACIDS. 

3. The precipitate produced by adding an excess of 
HO, S O3 to a solution of Ba O, N O5 weighed 5.942 
grammes. How much Ba 0, N O5 did the solution con- 
tain? 

172. 1st stage. S + O2={SO2,andK0,N05 + 2(ir0, ^Og) 

= KO,SO,. HO, SO, + MO, IVOs. 

2d stage. SO2 + HO, ]¥05 = HO, S 0, + ]\04. 
3d stage. 3]^04 + ccHO=2(HO,]\05) + J\02. 

r l¥02 + 02 = ]^®4• 
4th stage. < 2 SO, + 2 (H O, ]¥ O5) = 2 (^0, ^ 0,) 
( +2WO4. 

The last two stages are now repeated indefinitely, so 
long as there is a supply of sulphurous acid, oxygen, and 
steam, with the same amount of N O4 . 

1. How much sulphuric acid may be made by the above 
process from 100 kilogrammes of sulphur ? How many 
litres of acid having the Sp. Gr. 1.842 ? How many of 
acid of Sp. Gr. 1.734? How many cubic metres of oxy- 
gen must be used in the process ? How many cubic 
metres of air must pass through the lead chamber, sup- 
posing all its oxygen to be removed ? 

173. a. HO, SOs -\- Ati = H 0, S 0-\- Aq, 

1. How much water must one kilogramme of the mono- 
hydrated acid withdraw from the air in order to reduce its 
Sp. Gr. to 1.398 ? 

/. Na 0, CO^ -\-H0,S03-{-Aq = Na0,S0, + Aq 

^+C02. 

1. How much HO, S O3 is required to exactly neu- 
tralize 5.645 grammes of anhydrous carbonate of soda? 
How much acid of the Sp. Gr. 1.306 ? 



ACIDS. 69 

2. How much NaO, CO2 must be dissolved in one 
litre of water so as to make a solution such that one cubic 
centimetre will exactly neutralize 0.01 of a gramme of 
H 0, S O3 ? 

Yellow. White. 

^. Pb O + HO, SO^-\-Aq = PbO, S O3 + Aq. 

Black. Blue Solution. 

h. QxxQ -^ H 0, S 0^^ Aq = 0u0,8 0^-\- Aq, 
which, when evaporated, gives crystals of the composition 

Blue Vitriol. 

CuO, SO3.5HO. 

1. How much sulphate of lead can be made from twenty 
grammes of litharge ? How much sulphuric acid must be 
used in the process ? 

2. How much crystallized Blue Vitriol can be made 
from one kilogramme of oxide of copper ? How much 
sulphui'ic acid of Sp. Gr. 1.615 must be measured out for 
the process ? 

174. Cu + 2 {HO, SOs) = CuO, SO3 + 2 HO + SOa- 

175. C-\-2(HO,SOs) = 2HO + 2S02 + C02. 
WaO, GO,-\-SO^-}-Aq = JVaO,SO^-\-Aq-{-€Oi. 

1. How much sulphurous acid can be made from 4.562 
grammes of sulphuric acid by means of copper ? How 
much by means of charcoal ? How much anhydrous sul- 
phate of copper would be formed in the first case ? How 
much carbonic acid in the second ? How much carbonate 
of soda will the sulphurous acid in the two examples neu- 
tralize ? What is the volume of S O2, and what the 
volume of C O2 evolved in the second case ? 

2. How much copper and how much sulphuric acid 
must be used to make one litre of sulphurous acid gas ? 



70 ACIDS. 

How much to make 500 grammes of anhydrous sulphite 
of soda? 

3. By burning sulphur in one litre of oxygen, how much 
S O2 gas is obtained ? What is the Sp. Gr. of S O2 ? 

Phosphorus and Oxygen, 
Phosphoric Acid (POs). 
176. P + 5 O = P O5 (white powder). 

Colorless Fluid. 

3P + 6 {HO, NO,) + Aq = ZHO, ,PQ,-\- ^^ 

+ 5]\02. 

For preparation of phosphoric acid from bones, see § 144. 

At the ordinary temperature. 

3CaO, ePOs + 2(110, SOs)-\-aq= 2(CaO, S O3) 
+ [2^0, Ca 0] ,P O5 + aq. 

Intensely heated. 

2(CaO, S03)+ [2H0, CaOjePOs = SCaOePOs 

4-2(HO, SO3). 



Before ijcnition. 



3 HO, P O5 + 3 ilNH,-] 0,II0)-\-3 (Ag 0,NOs) + Aq 

Yellow. 

= 3AgO, ,P05 + 3 ([i^^4] 0, NO,) + Aq. 

After ignition. 

HO,PO,-^ INH,-] 0, HO + AgO, NO,-\-Aq = 

White. 

Ag 0, ,P O5 + [^^J 0,N0, + Aq. 

3H0, ,P0, + 2{MgO, SO,) + 3 [iV^^^] 0, HO 
-\-Aq == [N H4] 0, 2 Mg 0, eP O5 . 12 H O + 
2(lNH,^0,S0s)+Aq. 



ACIDS. 71 

[N H4] 0, 2 Mg 0, eP Og . 12 H when ignited resolves 
into 2 Mg O, bP O5 + W H3 + 13 H O. 

1. How much P Og and how much 3 H 0, cP Og can be 
obtained from 16 grammes of phosphorus ? 

2. By boiling one gramme of phosphorus in nitric acid 
until it dissolves, diluting, neutralizing with aqua ammonia, 
and precipitating with a solution of sulphate of magnesia, 
collecting and igniting the precipitate, how much wiU it be 
found to weigh ? 

3. How much nitrate of silver is required to precipitate 
the phosphoric acid made from one gramme of phos- 
phorus before ignition ? How much after ignition ? 

Cyanogen and Oxygen. 

179. Cy O = Cyanic Acid, which is mo7iobasic. 
Cy2 O2 = Fulminic Acid, which is hihasic. 
Cys O3 = Cyanuric Acid, which is trihasic. 

Boron and Oxygen. 
Boracic Acid (B O3). 

180. NaO, 2BO3 + HCl + Aq =. 2 {H 0, B 0^ 

+ iVa CT + Aq. 

The crystallized boracic acid is H O, B O3 . 2 H O. 
Dried at 100° it becomes H O, 2 B O3 . 2 H O. 

At a red heat it loses its water and melts, and on cooling 
it hardens to a vitreous mass. 



72 ACIDS. 

Silicon and Oxygen. 
Silicic Acid (Si O3). 
Na 0, Si O3 -{-HCl +Aq = HO, Si 0^ -\- Na 01 

-\-Aq. 

If the quantity of water is large, the hydrated silicic 
acid remains in solution. If the amount of water is small, 
it separates as gelatinous precipitate. 



SECOND GROUP : HYDROGEN ACIDS, OR COMPOUNDS OF 
THE HALOGENS WITH HYDROGEN. 

Chlorine and Hydrogen. 
Chlorohydric Acid (H CI). 

185. Na CI + HO, S Os = m 0, SO^-^ EL CI; or 

NaCl + 2 (HO, SO,) = Na 0, S 0, . HO, SO, 
+ HC1. 

Only one equivalent 0^ HO, S 0, is necessary to de- 
compose one equivalent of salt ; but then the last half of 
the H CI can be driven off only at a temperature suffi- 
ciently high to melt glass, so that with these proportions 
the process cannot be conducted in glass vessels. If two 
equivalents of HO, S 0, are used, the whole of the H CI 
is expeUed at a moderate temperature, and the process can 
then be conducted to its end in a glass flask or retort. 
Hence, in the manufactories, where the acid is generally 
generated in iron retorts, only one equivalent of HO, S O3 
is used, while in the laboratory, where glass vessels are 
employed in the process, two equivalents are taken to 



ACIDS. 73 

each equivalent of salt. The last we wiU assume to be 
the case in the following problems. 

1. How much chlorohydric acid gas can be made from 
4.562 grammes; from 25 kilogrammes; from 34.567 
grammes of common salt ? 

2. How much sulphuric acid is required to decompose 
the above amounts of common salt ? and how much bisul- 
phate of soda is in each case formed ? 

3. How many cubic centimetres of H CI can be made 
from 1 gramme ; from 5,643 grammes of Na CI ? 

4. How much salt and how much sulphuric acid are 
required in order to make one kilogramme, to make 
5.463 grammes, and to make one litre, of ffl CI ? 

5. How much H CI is contained in one litre of the 
liquid acid of Sp. Gr. 1.16, of Sp. Gr. 1.17, of Sp. Gr. 
1.14? 

6. How many cubic centimetres of gas are dissolved in 
one litre of the liquid acid of the above strengths ? 

7. How much Na CI, and how much HO, SO^, and how 
much water in the receiver, are required to make, — 

a. 20 kilogrammes of liquid acid of Sp. Gr. 1.13 ? 
h. 560.4 grammes of liquid acid of Sp. Gr. 1.18 ? 
c. 4 litres of liquid acid of Sp. Gr. 1.16 ? 

H + CI = M CI. 

1. One litre of hydrogen gas combines with what vol- 
ume of chlorine gas ? and what is the volume of hydro- 
chloric acid gas formed ? 

186. a.YQ^HCl-\-Aq = Fe 01 -{-Aq^ H. 

1. How much liquid acid, by weight and by measure, 
of Sp. Gr. 1,16 is required to dissolve 250 grammes of 

7 



74 ACIDS. 

iron ? How much chloride of iron would be obtained, 
and how much hydrogen gas, by measure, evolved ? 

^n -^ H CI -\- Aq = Sn Gl ^ Aq-^ H. 

1. Solve the last problem, substituting tin for iron. 

h. FegOs, 3 HO H- 3 ^C; + ^^ = Fe^ Ck + Aq. 

c. 2Fem-\- Cl^Aq= Fe^ Ck + Aq. 

d.NaO,G02-\-HCl-\-Aq = NaCl-\-Aq-\-€Oi. 

e. AgO,NO^-^ HCl-\-Aq = AgC\-\- HO, NOs-^Aq. 

1. How much liquid acid of Sp. Gr. 1.16 is required 
to dissolve 4 grammes of iron-rust? 

2. How many cubic centimetres of chlorine are required 
to convert one gramme of protochloride of iron into ses- 
quichloride ? 

3. How much liquid acid of Sp. Gr. 1.13 is required 
to neutralize one gramme of carbonate of soda ? 

4. How much H CI do 50 c. c. of a liquid contain 
which is exactly neutralized by one gramme of anhydrous 
carbonate of soda ? 

5. How much H CI do 50 c. c. of a liquid contain 
which gives, with an excess of nitrate of silver, a precipi- 
tate weighing 5.643 grammes ? 

Aqua Eegia. 

H0,N0,-\-ZHCl-\-aq = N02Ck-^Cl^ aq. 
Au + {N 0^ Ck -\-0l-\- aq) = Au Ck + aq + N O^. 



ACIDS. 75 

Bromohydric Acid (HBr). lodohydric Acid (HI). 

PBrs + SiTO = P03 + 3HBr. 
PI3 _^ 3 iTO = PO3 + 3 H I. 

Hydrofluoric Acid (H Fl). 

CaFl 4- RO,SO-\- aq = CaO, SO3 + HFl-\-aq. 
Si03 + 3HFl = 3HO + SiFl3. 

Tartaric Acid (2 H 0, Cs H4 0,o) • 

194. 2 UN H,-] 0, HO) + ^ H 0, O, H, 0,, -^ Aq =^ 
^INH,-] 0, GsJT,0,o + Aq. 

2 (K 0, CO,) + 2 ^ 0, Gs IT, O.o -}- Aq = 
2K0, Gsffi 0,0 + ^? + 2 C O2. 

2K0, GsJI, 0,0 + HGl + ^^ = HO, KO, Cs H4O10* 

-^KGl-\-Aq. 

2 (CaO, 00 + 2 (irO, KO, Gs H, 0,,) -\- Aq = 
2 CaO, C4 HsO.o + 2K0,GsH, 0,, + ^^ + 2 CO2. 

2/f 0, Gs H^ 6>,o -^2GaGl^Aq=2 Ca 0, Cg H4 0,o 

^2KGl^Aq. 

2 CaO, C4 HsOio + 2 {HO, S0-^^Aq==2 (Ca 0, S O3) 
^2 HO, G,HsO,,^Aq. 

* This salt is not absolutelj^ insoluble, but only difficultly soluble in -water, 
and hence is not completelj'- deposited in this reaction. 



76 ACIDS. 

1. How much KO, CO2 is required to exactly neu- 
tralize 5.462 grammes of tartaric acid ? 

2. How much H CI is required to convert 4.678 grammes 
of 2 K O, Cs H, 0,0 into H O, K O, C3 H4 Oio ? 

3. From ten kilogrammes of cream of tartar how much 
tartaric acid can be made ? 

OxaKc Acid (H 0, C2O3). 

The above is the sypabol of the acid dried at 100°. 
When crystallized, it contains two more equivalents of 
water, and corresponds to the formula H 0, C2 O3 . 2 H O. 

196. H O, Ca O3 . 2 ^0 + ^ {HO, SOs) = x {HO, SO,) 

197. h.KO,CO.,-{- HO, G^ 0, + Aq = KO, G^ 0, + Aq 

+ C O2. 

KO, 0, O3 + HO, C^O, + Aq = KO,^ a 6)3 + Aq. 

d. Ca 0, SO, ^ HO, Oi 0, -\- Aq = Ca O, C2O3 

-\-HO,SO, + Aq. 

Ca 0,SOs^ iNH,-] 0, G, 0, ^ Aq = Ca 0, C2 O3 
+ {NH,) 0,S0,-^ Aq. 

1. How much C O2 and how much C O will be ob- 
tained by decomposing five grammes of crystallized oxalic 
acid by sulphuric acid ? How many cubic centimetres of 
each gas ? 

2. How much crystallized oxalic acid must be used to 
yield one litre of C O2 and one litre of C O ? 

3. How much crystallized oxalic acid will exactly neu- 
tralize 1.456 grammes of carbonate of potassa ? 



ACIDS. 77 

Acetic Acid {HO, [C4 ^3] O3). 
198. PbO + HO, \_G^H,-] 0^ -\- Aq = Pb 0, [O.Hs] 0, 

■^Aq. 

CrystaUized acetate of lead = Pb O, [Q H3] O3 . 3 H 0. 

PhO, (GiHs) Os + HO, SOs + Aq = PbO, SO3 
-\-HO,lC,H,-] 0,-\-Aq. 



7* 



LIGHT METALS. 



FIKST GROUP : ALKALI METALS. 

Potassium (K). 

202. KO, aO^-\-HO,iG,H,-] 0,-]- Aq = KO,\_G,H,-\ 0, 

+ Jl^ 4- C 02. 

KO, CO2+ HO, S03 + ^^=XC>, ^^(93 + ^g+COj. 

1. How much HO, S O3 must be dissolved in water 
in order to make a litre of test acid such that one cubic 
centimetre will exactly neutralize one decigramme of 
KO, CO2? 

2. How much crystallized oxalic acid must be dissolved 
in water in order to make a litre of test acid such that 
one hundred cubic centimetres will exactly neutralize 6.92 
grammes of K O, C O2 ? 

203. K 0, G O2 -\- Ga 0, H -\- Aq = Ca O, C O2 

-^KO, HO -\- Aq. 

Melted together. 

204. d.SiOs-\-KO,I£0= K 0, Si O3 + H ©. 

Light Blue. 

e. Gic 0, S O3 -\- K 0, If -\- Aq === Cu O, H 

+ KO,SO, + Aq. 



LIGHT METALS, 79 

Intensely heated. 

205. KO, C O2 -(- 2 C = K + 3 C O. 

206. KO, 00^ + 2 {HO, SO^) -{-aq = KO, SO3. HO, S O3 

+ a^ + C02. 

207. KO, G0,-\- HO, N 0^ -\- aq = KO, N O5 -{- aq 

+ CO2. 

When heated. 

a. KO, NO5 = KO, NO3 + O2. 

J. K 0, NOg + 3 C + S = K S + 3 € O2 + TV. 

1. How many cubic centimetres of mixed gases are 
formed by the burning of one kilogramme of gunpowder, 
when measured at the standard temperature and pressure ? 
Assuming that the temperature at the time is 1000°, what 
would be the volume of the gases the moment after the 
explosion. 

2. Assuming that gunpowder occupies the same bulk as 
an equal weight of water, into how many times its own 
volume does it expand on burning ? Calculate both for 
0° and for 1000°. 

3. Assuming that the temperature, the moment after 
explosion, is 1000°, what would be the pressure on the in- 
terior surface of a bomb of 20 centimetres internal diame- 
ter when exploded filled with gunpowder ? 

208. a. K O, CI O5 = K CI + 6 O. 

1. How much does the oxygen contained in chlorate of 
potassa expand when the salt is decomposed ? Sp, Gr. 
of chlorate of potassa is 2 nearly. 

2. How much mechanical force would be required to 
reduce oxygen gas to the same degree of condensation in 
which it exists in the salt ? 

c. 3 (KO, CI 0,) + 4 {HO, ^ O3) = 2 (K 0, SO3 . HO, SO3) 

+ K0, C10; + 2ClO4. 



80 LIGHT METALS. 

f. -KO, C\Os+ ^ H Gl -^r Aq == KGl -{- Aq -\- 2 CI. 
6 {KO, HO) -\- <o CI -^ Aq = -KO, QIO, -Y 6 K 01 

+ Aq. 

210. KI ^ MnO, + 2{H0, S 0^) -\- aq = K 0, S Os 

-^ Mn 0, S Os -\- aq -{- I. 

211. H O, K 0, Cs H4 Oio = Cream of Tartar. 
Na 0, K O, Cs H4 Oio = Eochelle Salts. 

[N Hi] 0, K 0,t)8 Hi Oio = Ammoniated Tartar. 

Fea O3, K 0, Cg H4 Oio = Tartarized Iron 

Sbij O3, K 0, Cs H4 Oio = Tartar Emetic. 

B O3 K 0, Cs Hi Oio = Soluble Cream of Tartar. 

2 13. 4 (K 0, C O2) + 1 6 S = K 0, S O3 + 3 K Ss + 4 C ©8 . 

3(K0, C02) + 8S = KO, S2O2+2KS34-3CO2. 

The first or the last of these reactions takes place, 
according to the proportion of sulphur and the degree 
of temperature to wliich the mixture is exposed. If the 
sulphur is in excess, and the temperature of the mass 
raised to a red heat, pentasulphuret of potassium and sul- 
phate of potassa are formed. If, however, the sulphur is 
present in smaller quantity, and the temperature of the 
mass is not raised above its melting point, a mixture of 
tersulphide of potassium and hyposulphite of potassa re- 
sults. At the higher temperature, hyposulphite of potassa, 
which is always first formed, splits up into sulphate of 
potassa and pentasulphide of potassium. Thus, 

4 (K 0, S2 O2) = 3 K O, S O3 + K Ss . 

Either of these sulphides is decomposed by dilute acids, 
forming sulphide of hydrogen, and setting free as many 
equivalents of sulphur, less one, as are contained in the 
compound. 



LIGHT METALS. 81 

KO, S O3 + 4 C = K S + 4 C O. 

Sodium (Na). 

215. Problems on Common Salt. 

1. One gramme of salt contains how much chlorine and 
how much sodium ? 

2. One cubic centimetre of common salt, Sp. Gr. = 
2.13, contains how many cubic centimetres of sodium, and 
how many of chlorine gas ? 

218. Na CI + ^(9, /S O3 = Na 0, SO3 + H CI. 

219. NaO,S 03 + 40 = NaS + 4CO. 

220. Na S + CaO, C O2 = Ca S + Na O, C Oj. 

3 (NaO, S O3) + 13 C + 4 (CaO, CO^) = 3 NaO, COa') 
+ 3 Ca S, Ca + 14 C O. 

1. How much carbonate of soda can be made from 500 
kilogrammes of common salt ? How much sulphuric acid ? 
How much charcoal and how much chalk are required in 
the process, according to the theory ? 

2. If in a chemical process carbonate of potassa or car- 
bonate of soda may be used indifferently, which would be 
employed most profitably if the price were the same ? 

3. "What relation ought the price of crystallized car- 
bonate of soda to bear to that of the dry salt, if the intrin- 
sic value is alone considered ? 

221. Na 0, CO, -{- Ca 0, H -\- Aq = Ca 0, 0^ 

•\-NaO,HO.i-\-Aq. 



82 LIGHT METALS. 

222. Na 0, C Oa + 2 C = Na + 3 C O. 

223. 2 {Na 0, GO^)-\-n HO,PO, + Aq ^ 

HO, 2mO,POs-\-Aq^2€ O2. 

The symbol of crystallized phosphate of soda = 
HO, 2NaO, PO;.24HO. 

When heated. 

HO, 2NaO, POg.24HO=2NaO, PO5 + 25HO. 

3(Ag 0, NO,) + HO, 2 Na 0, P 0, -^ Aq = 

Yellow. 

3 KgO, V0,-{- 2 {Na0,N0s)-\~H0,N0s + Aq. 

While. 

2 {Ag 0, NO,) + 2iVa 0, P O5 + ^^ = 2 AgO, P O5 
-{- 2 {Na 0, N Os) + Aq. 

1. How much sodium can be made from one kilogramme 
of anhydrous carbonate of soda ? 

2. How much Ag O, N O5 is required to precipitate 
completely the phosphoric acid from 1.345 grammes of 
crystallized phosphate of soda ? How much, from the 
same amount of pyrophosphate of soda ? 

224. Na O, N O5 + 2 {HO, S 0,) = Na 0, S 0„ HO, SO, 

+ MO, ]\05. 

NaO,NO,^KGl-\-aq = '^2LQ\-\-KO,NO,-^aq. 

225. Na O, 2 B O3 . 10 H O = Symbol of common borax. 
NaO, 2BO3. 5H0= « " octohedral borax. 

1. "When soda nitre and potash nitre bear both the same 
price, from which can nitric acid be most profitably ex- 
tracted ? 

2. When potash nitre is worth 25 cents the kilogramme, 
and chloride of potassium 8 cents the kilogramme, at what 



LIGHT METALS. 83 

price of soda nitre will it be profitable to convert it into 
potash nitre, assuming that the cost of the process amounts 
to two cents on each kilogramme of potash nitre manu- 
factured ? 

3. What is the percentage composition of common 
borax ? What is that of octohedral borax ? 

4. How much borax glass can be made from 100 
grammes of common borax ? 

Ammonia ([N HJ 0). 

The above is the symbol of the base of the ammonia 
salts, and corresponds to K O and Na O. The student 
'must be careful not to confound it with ammonia gas, 
which has the symbol N H3 . 

227. K O, H + Fe = K O + Fe O + M. 

K O, N O5 -f 5 Fe = K O + 5 Fe O + J¥. 

3 (KO, H 0) -I- K 0, N O5 + 8 Fe = 4 K + 8 Fe O 

229. J\ H3 + II CI = [N HJ CI. 

[NHJCl + CaO, H0= Ca CI + 2HO + ]¥Il3. 

230. ^^ + M Ms = INff.l 0,H0^ 'Aq. 

231. INH,] 0,H0 + Aq-\-2U.^=lNH,-]S,HS-^Aq. 

The compound which remains in solution after pass- 
ing H S gas through aqua ammonia until saturation, is 
[N H4] S, H S, the sulphohydrate' of sulphide of ammo- 
nium, which is the reagent so much used in the laboratory, 
and incorrectly called sulphide of ammonium. On ex- 
posure to the air, the solution becomes yellow, owing to 



84 LIGHT METALS. 

the formation of bisulphide of ammonium, as is shown in 
the reaction below. 

\_NH,-\ S, HS + [_NH,-\ 0,H0-{- Aq = 
2iNH^-]S-\-Aq. 

Colorless Solution. 

2 (IN H,-] S, H S) + Aq -]- ^ O = 

Yellow Solution. 

lNH,-\ S, + IFH,-] 0, S, 0, + Aq. ' 

232. 2 [N H4] CI + 3 (Ca O, C 0^) = 3 Ca CI + JV H3 
+ 2 []V M J ®, 3 C O2 + H O. 

1. How many cubic centimetres do 17 grammes of 
ammonia gas occupy? How many do 36.5 grammes of 
chlorohydric acid gas occupy ? "When the two gases com- 
bine, in what proportions, by volume, do they unite ? How 
great is the condensation which results ? Sp. Gr. of 
[N H4] Ci = 1.5. 

2. How much ammonia gas can be obtained from 5 
grammes of chloride of ammonium. How much chloride 
of ammonium and how much Ca O must be used in order 
to prepare one litre of ammonia gas ? 

3. How much chloride of ammonium and how much 
lime must be used in order, to prepare one litre of aqua 
ammonia of Sp. Gr. = 0.9 ? How much water must be 
placed in the receiver ? 

4. How much ammonia gas is held in solution by one 
litre of aqua ammonia of Sp. Gr. 0.9 ? To how much 
[N H43 O, H O does this amount of gas correspond ? 

5. How much H 0, S O3 is required to neutralize four 
cubic centimetres t»f aqua ammonia of Sp. Gr. 0.9 ? 

6. How much H S gas will be absorbed by one litre 
of aqua ammonia of Sp. Gr. 0.9, assuming that only so 
much is taken up as is necessary to form the compound 



LIGHT METALS. 85 

[N H4] S, H S ? How much H 0, S O3 and Fe S will be 
required to produce this amount of gas ? How much 
additional aqua ammonia must be added to the above 
solution in order to change it to a solution of proto- 
sulphide of ammonium ? 



SECOND GROUP : THE ALKALINE EARTHS. 

Calcium (Ca). 

241. CaO,SO^-{-BaCl^Aq=B2^0,^03-\-CaCl-\-Aq. 

Ga 0, S 0, -\- H 0, G^ 0^ + ^^ = Ca 0, C^ O3 
-]-HO,SO,-\-Aq. 

242. HO,2N'aO,PO,-\-2GaGl-\-Aq=ILO,2Q2iO,VO, 

+ 2 iVa a + Aq. 

244. 2 (Ca 0, H O) + 2 CI = Ca 0, CI O + Ca CI. 
246. Ca O, C O2 + ^CT + ^^ = Ga Gl -\- Aq -{- CO2. 

1. How much Ca O can be obtained from 100 kilo- 
grammes of carbonate of lime ? How much Ca 0, H 
can be obtained from the same amount ? 

2. How much carbonate of lime must be burnt in order 
to yield 140 kilogrammes of quicklime ? How much to 
yield 185 kilogrammes of Ca O, H O ? How many cubic 
metres of C 0^ would be set free during the process ? 

3. How many cubic metres of C O2 can be absorbed by 
a quantity of milk of lime containing 5 kilogrammes of 
CaO? 

4. "What is the percentage composition of unburnt and 
of burnt gypsum ? 



86 LIGHT METALS. 

5. What is the percentage composition of phosphate of 
lime (3 Ca 0, P O5) ? 

6. How much chlorine and how much lime are re- 
quired to make 100 kilogrammes of chloride of lime, 
assuming that its composition is expressed by the symbol 
Ca O, CI O + Ca CI ? How much Mn O2 and how much 
muriatic acid of Sp. Gr. 1.15 will yield the requisite 
amount of chlorine ? 

7. To how much Ca O, how much Ca CI, and how much 
Ca O, S O3 do 2.5 grammes of carbonate of lime correspond? 

Barium and Strontium (Ba and Sr). 

248. Ba O, S O3 + 4 C = Ba S + 4 C €». 

Ba S -{- H CI -^ Aq = Ba CI -\- Aq -\-m^. 

QnO + Ba S -\- Aq = CnS -\- Ba 0, H -\- Aq. 

Ba CI -\- NaO, S 0, -\- Aq = Ba 0, S O3 + Na CI 
+ Aq. 

Sr O, S O3 + 4 C = Sr S + 4 C ©. 

SrS-^HO, NO, J\-Aq = SrO,NO,-\-Aq-\-m^. 

Sr 0, N 0, + Na 0, S 0, + Aq = Sr 0, S O3 
NaO,NOs-\-Aq. 

1. What is the percentage composition of sulphate of 
baryta ? 

2. How much chloride of barium can be made from 5 
kilogrammes of sulphate of baryta ? How much nitrate 
of baryta, and how much Ba O, can be made from the 
same amount of sulphate of baryta ? 

3. To how much sulphuric acid do 4.567 grammes of 
sulphate of baryta correspond ? To how much chloride 
of barium do they correspond ? 



LIGHT METALS. 87 

4. How much nitrate of str-ontia can be made from one 
kilogramme of sulphate of strontia ? How much carbon 
and how much nitric acid of Sp. Gr. 1.4 is required in 
the process ? 

Magnesium (Mg). 

249. The symbol of serpentine mineral is 

9[Mg,Fe]0, 4Si03, 6H0. 

The symbol [Mg, Fe] O indicates that a portion of the 
magnesium iu the base is replaced by oxide of iron, and 
the whole stands for but one equivalent of base. We 
frequently write, instead of the portion enclosed in brack- 
ets, the general symbol R, when the symbol of the 
mixed base becomes E. O, and that of serpentine mineral 
9 E O, 4 Si O3. It must be noticed, that Mg and Fe when 
enclosed in brackets, with a comma between, as above, no 
longer stand for an equivalent of each metal. On the 
other hand, the two together make but one equivalent of 
metal, represented by R in the other mode of writing. 
Moreover, nothing is intended to be indicated in regard to 
the relative proportions of the two metals mixed together 
to form one equivalent, as they vary in different speci- 
mens of the same mineral. This method of writing the 
symbols of compounds containing isomorphous constitu- 
ents Avill be constantly used hereafter. 

9 [Mg, Fe] 0, 4 Si O3 . 6 H O + 9 (^6>, ^ 0^) -\- Aq ^ 
4 Si O3 + 9 [j%, Fe-] 0,S0,-{- Aq. 

Since sulphate of magnesia and sulphate of protoxide 
of iron have the same crystalline form, we shall obtain, on 
evaporating the solution, crystals containing both salts. 
We can prevent the sulphate of protoxide of iron from 



88 LIGHT METALS. 

crystallizing, by converting it into sulphate of sesquioxide 
of iron by means of nitric acid. Thus, 

6 {Fe 0, aS' O3) + 3 {HO, S 0,) + HO, NO, -^ Aq ^ 
3 {Fe^ O3, 3 ^ O3) + ^^ + W ©2. 

Talc = 6MgO, 5 SiOs, 2 H 0. 

Meerschaum = Mg O, Si O3, H 0. 

Hornblende = 4 [Mg, Ca, Fe] 0, 3 Si O3. 

Augite = 3 [Ca, Fe, Mg] 0, 2 Si O3. 

250. 5 {Mg 0, SO,) + 5 {KO, 00^) -{-Aq=d (MgO, CO2 . aq) 

+ MgO, HO + % 0, 2 (76>2 + 5 {K 0, S 0^) -^ Aq. 

The relative proportions of carbonate of magnesia and 
of hydrate of magnesia vary with the temperature and 
other circumstances attending the precipitation. 

251. M.g0,Q0^-\-HGl^Aq = Mga^Aq-\-i^O2. 

2 {Mg 0, S0s)-\-H0,2 Na 0,P0,-\- {N H,-\ 0,H0 
-{-Aq= [NHJ 0, 2MgO, PO5 + 2 {Na 0, S 0,) 

JrAq. 

When heated. 

(NH4)0,2MgO,P05=2MgO,P054-WM3 + HO. 

1. What is the percentage composition of talc ? What 
that of hornblende and augite, assuming that the whole of 
the base in either case is Mg O ? 

2. From a solution of sulphate of magnesia the whole 
of the magnesia was precipitated by phosphate of soda 
and ammonia. This precipitate, after ignition, was found 
to weigh 2.456 grammes. How much sulphate of mag- 
nesia was contained in the solution ? 



LIGHT METALS. 89 

Aluminum (Al). 

258. AI2O3, Si O3 . 2 H O + 3 {HO, S O3) -\- aq = SiOs 

+ AI2O3, 3S03 + a(?. 

AU 0^,^SO^ + Z{N'aO, G 0^)^Aq = A\0^,ZB.O 
+ 3 {Na 0, S03)+Aq-\-C ©2. 

AI2O3, 3 HO + KO, HO-\-Aq = KO, Ak O3 + Aq. 

KO,SO^. Ak O3, 3 aS C>3 + 3 {Fa 0, O 0,) -{- Aq = 
AI2 O3, 3 HO + 3 (iVa O^S O3) +KO,SOs-{- Aq 

+ C®2. . 

Al, Os, 3S0, -\- S (Pb 0, 10, H,-\ O3) + Aq = 
3(PbO, SO3) +^4 0„ 3 lO,H,-\ 0, + Aq. 

Symbols of Isomorphous Alums. 

Potassa, Alumina, Alum. 
K 0, S O3 . AI2 O3, 3 S O3 . 24 H O. 

Soda, Alumina, Alum. 
NaO, S O3 . AI2O3, 3 S O3 . 24 HO. 

Ammonia, Alumina, Alum. 
[N H4] O, SO3 . AI3 O3, 3 SO3 . 24HO. 

Potassa, Chrome, Alum. 
KO, S O3 . Cr2 O3, 3 S O3 . 24 HO. 

Soda, Chrome, Alum. 
Na 0, S O3 . Crg O3, 3 S O3 . 24 H O. 

Ammonia, Chrome, Alum. 
[N HJ 0, S O3 . Cr2 O3, 3 S O3 . 24HO. 



90 LIGHT METALS. 

Potassa, Iron, Alum. 
K O, S O3 . Fea O3, 3 S O3 . 24 HO. 

Soda, Iron, Alum. 
NaO, S O3 . Fes O3, 3 S O3 . 24:HO. 

, Ammonia, Iron, Alum. 
[NH4] 0, S O3 . FeaOs, 3 S O3 . 24HO. 

Symbols of the most important Silicates of Alumina. 

2AI2O3, SiOg, Staurotide. 

3 AI2O3, 2 SiOs, Andalusite. 

3 AI2 O3 , 2 Si O3 , Ky anite. 

3 AI2 O3, 2 Si [O, Fl]3, Topaz, 

K O, Si O3 . AI2 O3, 3 Si O3, Common Felspar. 

Na O, Si O3 . AI2 O3, 3 Si O3, Albite. 

[Ca, Na] 0, Si O3 . AI2 O3, Si O3, Labradorite. 

K 0, Si O3 . 4 (AI2 O3, Si O3), Common Mica. 

3 R O,* Si O3 . R. 03,t Si O3, Garnet. 

1. What is the percentage composition of staurotide ? 
What is that of kyanite ? 

2. An analysis of one of the above silicates would give 
the foUowing percentage composition. 





Silica, 




64.76 




Potassa, 




16.87 




Alumina, 




18.37 
100.00 


What is 


the symbol 


of the 


mineral ? 



* K = Fe 0, Mn 0, Mg 0, or Ca 0. 
t Ks Os = AI2 O3, Fe2, O3, or Cra O3. 



LIGHT METALS. 91 



Solution. — This problem is evidently the reverse of deducing the per- 
centage composition from the symbol ; but it does not admit, like that, of 
a definite solution, for while there is but one percentage composition 
corresponding to a given symbol, there may be an infinite number of 
symbols corresponding to a given percentage composition. This can 
easily be made clear by an example. The commonly received symbol 
of alcohol is [C4 H5] 0, H = C4 Ho O2 . The percentage composition 
is easily ascertained. Thus, 

C4 Hg O2 

24 + 6 + 16 = 46. 
46 : 24 = 100 : a; = 52.18 per cent of carbon. 
46 : 6 = 100 : x = 13.04 per cent of hydrogen. 
46 : 16 = 100 : x = 34.78 per cent of oxygen. 

Percent. 

Carbon, 52.18 = Ca = 12 or C4 = 24 or Cs = 36 

Hydrogen, 13.04 = H3 = 3 " Hs = 6 " H9 = 9 

Oxygen, 34.78 = =_8 " O2 =_16 " O3 = 2i 

100.00 23 46 69 

This percentage composition evidently corresponds not only to C4 Hs O2 , 
but also to C2 H3 0, to Cg H9 O3 , and to any other symbol -which is a 
multiple of the first; for, taking the per cent of carbon as an example, 
■we have 

100 : 52.18 = 23 : 12 = 46 : 24 = 69 : 36 = 92 : 48, &c. 

If, then, we had given the percentage composition of alcohol, it would 
be impossible to determine, without other data, whether the symbol was 
C2 Ha 0, or some multiple of it. If, however, we had also given that the 
sum of the equivalents of the elements of alcohol equalled 46, then we 
could easily reverse the above process. Thus, 

100 : 52.18 = 46 : a; = 24, the sum of the equivalents of carbon. 
100 : 13.04 = 46 : a; = 6, " " " " hydrogen. 

100 : 34.78 = 46 : a; = 16, " " " « oxygen. 

— = 4, number of equivalents of carbon. 



hydrogen. 



— == 6, " " " 

16 

— = 2, " " " " oxygen. 

Assuming, however, that we had no means of ascertaining the sum of 
the equivalents in alcohol, then, although we could not definitely fix its 
sjmibol, yet nevertheless we could easily find which of all the possible 
symbols expressed its composition in the simplest terms ; in other words, 



92 LIGHT METALS. 



■with the fewest number of whole equivalents. For this purpose, assume 
for a moment that the sum of the equivalents is equal to 100, then 

52.18 = the sum of the equivalents of carbon. 
13.04= " " " " hydrogen. 

34.78 = " " " " oxygen. 

— ^ = 8.697, number of equivalents of carbon. 

13^ = 13.04, « " " " hydrogen. 

84 78 

— ^ = 4.348, " " " " oxygen. 

8 

These are the number of equivalents of each element on the supposition 
that the sum of the equivalents in alcohol is equal to 100. Any other 
possible number of equivalents must be either a multiple or a submul- 
tiple of these, and we can easily find the fewest number of whole 
equivalents possible, by seeking for the three smallest whole numbers 
■which stand to each other in the relation of 8.697 : 13.08 : 4.348, 

■which will be found to be 2:3:1. 

Hence the simplest possible symbol is C2 H3 0, but, from anything we are 
assumed to know, the symbol may be any multiple of this ; and for con- 
siderations which cannot be discussed in this connection, chemists usually 
assign to alcohol the symbol C4 He O2 , which is double the above. 

The symbol thus obtained expresses merely the relative number of 
equivalents of each element present in the compound, and gives no in- 
formation in regard to the grouping of the elements. Such symbols are 
called empirical symbols, to distinguish them from the rational symbols, 
which indicate the manner in which the elements are supposed to be 
arranged. The rational symbol of alcohol is [C4 H5] 0, H 0. This in- 
dicates not only that alcohol consists of four equivalents of carbon, six 
of hydrogen, and two of oxygen, but also that it is the hydrated oxide 
of a compound radical called ethyle. It must be carefully noticed, how- 
ever, that the empirical symbols fully represent all our positive knowl- 
edge. They alone are not hable to be changed. The gi-ouping of 
elements in a compound is a matter of theory, and the rational symbols 
are liable to constant changes, as the opinions of chemists on this sub- 
ject vary. 

From the example just discussed we can easily deduce the following 
rule for finding the empirical symbol of a compound from its percentage 
composition. Divide the per cent of each element entering into the com- 
poundhy its chemical equivalent^ and find the simplest series ofichole num- 
bers to ivMch these results correspond. To apply this rule to the problem 
under consideration. 



LIGHT METALS. 93 



— 1-— = 1.43, number of equivalents of silica. 
45.3 

—1— = 0.3575, " " " " potassa. 
47.2 

i-— - = 0.3575, " " " " alumina. 
51.4 

143 : 0.3575 : 0.3575 = 4:1:1. 

Empirical symbol, AI2 O3 , K 0, 4 Si O3 . 

Rational symbol, K 0, Si O3 , AI2 O3 , 8 Si Oa . 

3. An analysis of one of the silicates of alumina would 
give the following percentage composition. 

Per cent. 

Silica, 53.29 

Lime, 16.47 

Alumina, 30.24 

100.00 

What is the symbol of the minei'al ? 

Ans. Ca 0, Si O3 . A\, O3 Si O3. 

Second Method of Solution. — By inspecting the formula obtained by 
solving the problem according to the method just described, the student 
will see that the amount of oxygen in the acid stands in a very simple 
relation to that in the bases. This relation is 1 : 3 : 6, corresponding to 
Ca 0, AI2 O3 , and 2 Si O3 . It has been shown in the text-book, § 200, 
that a similar simple ratio exists between the amount of oxygen in the 
acid and that in the bases of all oxygen salts. The I'atio can easUy be 
found from the percentage composition. For this purpose we have 
merely to calculate the amount of oxygen in the per cent of the acid and 
bases indicated by analysis, and find the simplest ratio in which these 
amounts stand to each other. In our example, 

53.29 per cent of silica contains 28.24 parts of oxygen. 
16.47 " " lime " 4.71 " " " 

30.24 " " alumina " 14.12 " " " 

According to the principle just stated, these numbers ought to stand to 
each other in some simple ratio, and it can easily be seen that 

4.71 : 14.12 : 28.24 = 1:3:6. 

From this ratio we can easily deduce the symbol, for one equivalent of 
oxygen corresponds to one equivalent of Ca 0, three equivalents of oxy- 
gen correspond to one equivalent of AI2 O3 , and sis of oxygen to two 



94 LIGHT METALS. 

equivalents of Si O3 . Hence the symbol is Ca 0, AI2 O3 , 2 Si O3 , which 
we may write as above, Ca 0, Si O3 . AI2 O3 , Si O3. For convenience 
in calculating the amount of oxygen from the per cent of acids or bases 
indicated by analysis, Table VII. has been added at the end of the book, 
which gives the per cent of oxygen, together with its logarithm con- 
tained in the bases and acids of most common occurrence. 

In deducing empirical symbols from the results of actual analysis, it 
must be remembered that our processes are not absolutely accurate, and 
that therefore we must not expect to find more than a close approxima- 
tion to a simple ratio between the oxj-gen in the base and that in the 
acid. Again, in mineral compounds, it is very frequently the case that 
isomorphous bases replace each other to a greater or less extent. This is 
the case in common garnet, the symbol of which may be written thus : 

8 [Fe, Jin, Mg, Ca] 0, Si O3 . [Al2Fe2] O3, Si O3. 
We generally, however, write the symbol as on page 90 : 

3 R 0, Si O3 . E2 O3 , Si O3 . 
Here E stands for the sum of all the protoxide bases, which make 
together but one equivalent of base, and E2 O3 for the sum of all the 
sesquioxide bases, which also make together but one equivalent of base. 
Such general symbols as these give all the information in regard to the 
constitution of the mineral which is required. In deducing such sym- 
bols, it is evident that the oxygen of all the protoxide bases must be 
added together to obtain the amount of oxygen in the assumed base 
E 0, and aU the oxygen of the sesquioxide bases must be added together 
in the same way in order to obtain the amount of oxygen in the assumed 
base E2 O3 . From these sums we can easily obtain the required oxygen 
ratio. 

4. An analysis of andesine (a mineral allied to felspar) 
yielded the following result. 











Proportion of Oxyg'en. 


Silicic Acid, 




59.60 


30.90 in Si 0. 


Alumina, 
Sesquioxide 


of Iron, 


24.28 
1.58 


":S}="-™""^o 


Lime, 




5.77 


1.61- 




Magnesia, 




1.08 


0.37 


- = 3.79 in RO. 


Soda, 




6.53 


1.65 




Potassa, 




1.08 


0.16. 





99.92 

What is the symbol of the mineral ? 



LIGHT METALS. 95 

Solution. — The ratio of the oxygen in E 0, R2 O3, and Si O3 is 
3.79 : 11.70 : 30.90 = 1 : 3.08 : 8.1, 
for which we may substitute, for reasons stated above, 
1:3:8. 
One equivalent of oxygen corresponds to R 0. • 

Three equivalents of oxygen correspond to Eg O3 . 
Eight equivalents of oxygen correspond to ^ Si O3 • 
But as we do not admit fractional equivalents, we may multiply the 
whole by three, when we obtain the empirical symbol 

3E0, 3E2 03,8Si03; 
from which we may deduce the rational symbol 
3 E 0, 2 Si O3 . 3 (E2 03, 2 Si O3). 

5. Deduce the symbols of the silicious minerals of which 
the following are analyses. 

1. 
Silicic Acid, 65.72 

Sesquioxide of Iron, 
Alumina, 18.57 

Lime, 0.34 

Magnesia, 0.10 

Potassa, 14.02 

Soda, 1.25 

100.00 100.0 100.04 100.16 



2. 

68.4 


3, 

44.12 


63.70 


0.1 


0.70 


0.50 


20.8 


35.12 


23.95 


0.2 


19.02 


2.05 




0.66 


0.65 




0.25 


1.20 


10.5 


0.27 


8.11 



HEAVY METALS. 



FIRST GROUP OF THE HEAVY METALS. 

Iron (Fe). 
286. a. 4 Fe + O = Fe4 O. 

b. 3 Fe4 O + 13 O = 4 (Fe 0, Fe^ O3). 

c. 2 (Fe 0, Fe.2 O3) + O = 3 Fe^ O3. 

d. 2 (Fe O, S O3. 6 H O) when heated resolves into 

Fg O3, S O3 + S O2 + 12 H O ; by further heat- 
ing, Fe^ O3, S O3 = Fea O3 + S O3. 

e. 3 Fe + 4 O = Fe 0,'Fe2 O3. 

/. 2 (Fe 0, Fea O3) H- 9 ^ + © = 3 (Fe^ oTs H 0). 
g. Fe O, Fes O3 -\- 2 G 0^ + Aq = Fe^ O3 
^Fe 0,2 O 0,-\-Aq. 

Red. 

2 (Fe 0, 2 O2) + ^^ + -O = Fe^ O3, 3 H 
^4,0 0,-^ Aq. 
285. Fe + IlO,S03-{-Aq = FeO,SO,-j-Aq-{-n^ 



HEAVY METALS. 97 

285. Fe + Cii 0, ^ O3 + ^^ = Cu + Fe 0, S 0, 

Red. 

a. 6 (Fe 0, S 0,) -\- Aq + 3 O = ¥e, O3, 3 H 

Yellow-Brown. 

-{-2(Fe^0s, 3 S 0,)+Aq. 
h. 6 {Fe O, S 0,) + 3{H0, S 0,) -{- H 0, N 0^ 

Yellow-Brown. 

■\-Aq=3 {Fe^ Os, 3 S O3) + .4^ + ]¥ O2 . 

Light Green. 

c. FeO,SOs-\-lIfJI,2 0,ffO-^Aq = FeO,B.O 

+ [iV^^4] 0, SOs-\-Aq. 
Fe^ Os, 3SOs + 3 ([JSTIf,] 0, H 0) -\- Aq = 

Red. 

Fe2 O3, 3 H + 3 ([J^IIi'] 0, S 0^) + Aq. 

d. '2^YQ-}-A:HO,NO,-^rAq = Fe^Os,3NO,-\-Aq 

288. \_H0, 2 Na 0] P Os + 3 {Fe 0, S 0^) -\- Aq = 

White. 

3 Fe O, V 0,^2Na 0, SO^^HO, SOs 

~\-Aq. 

[HO, 2NaO']POs-\- Fe^ Os, 3 S Os + Aq = 

While. 

Fe^Os ,P05.4HO+2 {M 0, SOs)-\-HO, SO^+Aq. 

290. 3 Fe O + 2 Fea O3 + ^ H Cy -^ Aq = 

Blue. 

3 Fe Cy, 2 Fca €73 + 9 HO + Aq. 

Blue. 

291. 3 Fe Cy, 2 Fcg Cyg + % {K 0, HO) + Aq = 

, Red. 

2 (Fe2 O3, 3 H 0) + 3 (2 ir (7y, Fe Cy) + Aq. 

292. a. 3 (2 ^ (7y, Fe Cy) + 2 (^eg Os, 3 SOs)+Aq = 

3 Fe Cy^^'Fea Cyg + 6 (.ff 0, >«? O3) + Aq,. 

9 



98 HEAVY METALS. 

292. 5. 2 ^ Gy, Fe Cy ^ 2 {Fe 0, S 0,) -^ Aq = 

White. 

3 Fe Cy 4- 2 Z" 0, 5 O3 + Aq. 

White. Blue. 

9 Fe Cy + 3 O = Fe, O3 + 3 Fe Cy, 2 Fez Cyg. 
c. 2 K Cy, Fe Cy -\- 2 {Cu 0, S O3) -\- Aq = 

Purple. 

2 Cu Cy, Fe Cy. Aq + 2 {KO, S 0,) + Aq. 
2 Z- (7y, Fe Cy -i- 2 (Pb 0, ]^ 0,) -\- Aq = 

White. 

2 Pb Cy, Fe Cy. Aq + 2 (JT 0, iV^Os) + Aq. 
A large number of similar compounds having the 

general symbol 2 R Cy, Fe Cy. Aq, may be 
formed thus : — 

White. Pale Yellow. 

2 H Cy, Fe Cy. Aq. 2 [N H/] Cy, Fe Cy. Aq. 

Yellow. Yellow. 

2 Na Cy, Fe Cy. Aq. 2 Ba Cy, Fe Cy. Aq. 

Pale Yellow. White. 

2 Mg Cy, Fe Cy. Aq. 2 Zn Cy, Fe Cy. Aq. 

There are also compounds in which the two equiva- 
lents of R in the general symbol are replaced by 
different metals thus : — 

White. Yellow. 

Yellow Salt. Red Salt. 

293. 2 (2 X Cy, Fe Cy)^ Cl-{- Aq =Z K Cy, Fe^ Cy, 

-^KCl-\- Aq. 

3 K Cy, Fe^ Cy, + 3 {Fe 0, S 0,) -\- Aq = 

3 Fe Cy,'"Fe2 Cys + 3 {K 0, S 0,) + Aq. 



HEAVY METALS. 99 

293. There also may be formed a large nmnber of 
similar compounds having the general symbol 
3 E. Cy, Fcg Cys, such as : — 

Brown. Lijht-Red. 

8 H Cy, Fe, Cy,. 8 Ca Oy, Fe^ Cyj. 

{^B^Cyl-'^'^Cys. Aq. 

Light-Gtecn Solution. Black. 

294 Fe 0, S 0, + \_N H,-\ ^ + ^^ = Fe S 
+ IN-H,-] 0,SO,-\-Aq. 

Black. Lig-ht-Green. 

295. Fe S + a^ + 4 O = J^e 0, ^ O3 + a^. 



Manganese (Mn). 

299. Mn O2 + iT 0, S 0^ -\- aq = Mn 0, S03-\-aq 

+ 0. 

Black. Lijht-Pink Solution. 

MnO^-^^ 2 H Gl -]- Aq = Mn 01 -^ Aq -\- CI. 

Light-Pink Solution. Brown. 

300. a. 6 {Mn 0, /S O3) + .ig + 3 O = Muj O3, 3 H O 

+ 2 {Mn^ Os, S S Os)+ Aq. 

Light-Pink Solution. 'White. 

b. MnO,S03-{-l]^iri']0,H:0-\-Aq = MnO,'H.O 

-\-[i^ir,-]o,so, + Aq. 

White. Brown. 

2 (Mn 0, H 0) + «? + O = Muj O3, 3 H O 
-\-aq. 

Lig-ht-Pink Solution. Flesh-colored. 

Mn 0, S Os -\- [iV^ ff,-] S -\- Aq = Mn S 
-{-[JSTH,-] 0,SO,-i-Aq. 

Melted together. 

301. MnO^-f K 0,11 0-^0= K0,Mn03 + UO. 



100 HEAVY METALS. 



Green Solution, 

301. 3 {K 0, Mn O3) + ^y + 2 C O2 = Mn O2 

Crimson Solution. 

-^KO, Mn^ Ot-\-2{KO,G 0^) + Aq. 

Green Solution. 

3 {K 0, Mn Os) -}- 2 (JIO, S O3) -{- Aq = 

Crimson Solution. 

Mn Oi+KO, Mn^ Ot^2{KO,S 0,)J^Aq. 



Cobalt (Co) + Nickel (Ni). 

Pink Solution. Blaclc. 

303. Co 0, S Os -{- [N- H,-\ S -[- Aq = Qo ^ 



\_NH,-] 0,SO, + Aq. 

Green Solution. 

+ [i\rir,] o,sOs + Aq. 



Green Solution. Black. 

M 0, S Os -^ [JSr H,-] S -{- Aq = m ^ 



Zinc (Zn). 

311. Zn -{- H 0, S 0^ -{- Aq = Za 0, S Os -\- Aq 

+ H. 

White. 

312. a. Zn 0, S Os -\- KO, H 0-{- Aq =^ZnO, no 

■\-KO,SOs-\- Aq. 
This precipitate dissolves in an excess of K 0, HO 
+ Aq. 

White. 

b. 2k 0, S Os -i- [li H,-] S -\- Aq = Zn ^ 
^iNH,-\0,SOs + Aq. 



HEAVY METALS. ^01, 

312. c. 5 {Zn 0, S 0,) -^ 6 {Na 0, G 0,) -{- Aq = 

While. White, 

2 (Zn 0, C Oa) + 3 (Zn O, H 0) 
+ 5 (iVa 0, S 0^) -\- Aq -\- 3 € O,. 
This, precipitate is a mixture of Zn O, C O2, and 
Zn O, H O, but in variable proportions. 



Yellow. 

315. Cd 0,S Os + JIS-\-Aq=CdS-\-irO,S03 



Dark-Brown. Gray-White. White. 

317. The oxides of tin are Sn O ; Sng O3 ; and Sn Oj. 



Cadmium (Cd). 

a 

+ Aq. 

Tin (Sn). 

Dark-Bi 

tin are Sn 

319. Sn -^ H Gl ^ aq = Sn 01 -\- aq -{- n. 

White. 

320. Sn Gl + \_NH^ 0, H -{- Aq = SnO, HO 

+ [iV^^4] Gl^Aq. 

321. Sn Gl -\- Gl -\- Aq = Sn Gl, + Aq. 

Sn -\- 2 I£ Gl -i- IT 0, W Os -\- Aq = Sn CI, 

-\-Aq + NOs. 

White. 

SnGk-\-2 ilJ^Ifi'] 0,irO)^Aq = lIO, Sn 0^ 

+ 2 [i^^,] Gl-\-Aq. 
SnO+KO,HO-^Aq = KO,Sn -\- Aq.. 
HO, Sn O2 + ^ 0, HO -\-Aq = KO, Sn Q, 

+ Aq. 

9* 



102 HEAVY METALS. 

By evaporation of the last solution we can obtain 

crystals of K O, Sn Oj. 4 H O. 

We may also prepare Na O, Sn Og . 4 H O. 

White. 

S24. 5 Sn + 10 {H 0, N 0,) -\- Aq = Sng 0,o. 10 H O 
^Aq -\-10NOi. 
Sng Oio. 10 H O + K 0, H -\- Aq = 

KOySn, 0,,^Aq. 
By evaporation we can obtain crystals of 

White. 

K 0, Sng Oio. 4 H O. 

Brown. 

325. Sn CI -{- ITS -{- Aq =^ Sxi S -}- H Gl -{- Aq. 

Yellow. 

Sn Ck + 2 HS-\-Aq = Sn ^^ -\- ^ H Gl -\- Aq. 



SECOND GROUP OF THE HEAVY METALS. 

Lead (Pb). 

Black. Red or Yellow. Red-Yellow. 

331. The oxides of lead are Pbj O ; Pb O ; Pba O3 ; 

Dark-BrowD, 

and Pb Oj. 
334, B Vh -\- 4. {H O, N 0,) -\- aq = B (Pb 0, W 0^) 

+ a^ + ]¥ O2. 
3 Pb 0-1- 3 (^0, iV^ O5) + a^ = 3 (Pb 0, N 0,) 
-\- aq. 



HEAVY METALS. 103 

White. 

335. Ph 0, N Os + HO, S 0^-\-Aq = Pb O, S O3 

+ HO,NOs + Aq. 
Ak O3, 3 >Sf (93 + 3 {Pb 0, [C; ^3] O3) +Aq = 

White. 

3 (Pb O, S O3) + Ak O3, 3 [Ci H,-] O3 + Aq. 

White. 

336. Pb O + ^ (7? + a^ = Pb CI + aq. 
Vh O + H Gl -\- Aq = Pb CI + Aq. 

337. Acetate of oxide of lead = Pb O, [C^ H3] O3 . 3 H O. 
Basic acetate of oxide of lead = 

3 PbO, [QHsjOa. HO. 

338. 2 (Pb 0, [ C4 ^3] 6>3) + 2 IT 0, 0^ B, 0,, -\-Aq^ 

White. 

2 Pb O, Cs H, Oio + 2 (IT (9, [ Oi H,-\ 0,).+ Aq. 

White. 

PbO,N'0,-\-lNH,-\ 0,HO-\-Aq = VhO,B.O 
JrWH,-]0, NO,-^Aq. 

339. Pb 0, [C, Bs] Os + Aq 4- 2 Pb O = 

3Pb 0,lG,ff,-] 0, + Aq. 
3 Pb 0, IC ffs] O3 + ^? + 2 C O2 = 

While. 

2 (Pb O, C O2) + P& 0, [a ^3] O3 + Aq. 

340. Zn -f P5 (9, [C; Hs] O3 -{- Aq == Pb 

+ ^« 6>, [en,-] Os-i-Aq. 

Blaclc. 

341. P^ 0, [O; 1^3] O3 + ^ .S' + ^^ = Pb s 

+ ^0, [(7,^3] 0,-\-Aq. 

342. Pb S + 3 O = Pb O + S O2, also Pb S 

White. 

+ 4O = Pb0, SO3. 



104 HEAYT METALS. 

342. By roasting galena we obtain a mixture of Pb O 
with a small amount of Pb O, S O3. By then 
melting together Pb O or Pb 0, S O3 and an 
excess of Pb S, we obtain metallic lead and sul- 
phurous acid, thus : — 
2 Pb O + Pb S = 3 Pb + S O2, also Pb 0, S O3 
+ PbS = 2Pb + 2S02. 

Bismuth (Bi). 

847. 2 Bi + 4 (^ (9, II O5) + «^ = Bi^ O3, 3 J^ 0^ 
-\-aq-\-N O2. 

White. 

4 (Bi, Oe, SI^Os) -\- Ag = 3 (Bi^ O3, N O5) 

+ Bh O3, 9 i\r O5 + Aq. 

Brownish-Black. 

Bh O3, d If Os -\- 3 II S -{- Aq = Bk S3 
-\- 9 (HO, If 0,) + Aq. 

Copper (Cu). 

Green. 

349. Malachite = Ou 0, HO. Cu O, CO2. 

Blue Carbonate of Copper = 

Blue. 

CuO, HO. 2 CuO, CO2. 

Red. Black. 

350. The oxides of copper are Cu2 and Cu O. 

Blue Solution. Blue. 

352. Cu 0, S 0, + K 0, H -\- Aq = Cvi O, n O 

■ -\-K0, SO,-\-Aq. 



HEAVY METALS. l05 

Blue. Black. 

352. By boiling, Cu 0, B. -{- Aq ^ Cn O -}- Aq. 

Light-Blue Solution. 

353. GuO, ^03 + 2 ilNH,-] 0, H 0) -^ Aq =^ 

Very deep Blue Solution. 

Oit 0, S Os- 2 IT Bs. H -i- Aq = 

[^^l] 0, S 0,. [_NH,-\ 0,H0^Aq,2. 

solution which, when treated with alcohol, yields 
crystals having the composition 

Deep Blue. 

JN^^I 0, SO3. [NHJO, HO. 
These, when heated, are resolved into 

Green Powder. 

|n H^3 I 0, S O3 + []¥ H4] O, H O. 

854. 2((7mO, >S03) + 2(^0, ir(9)+^g — 0* = 

Yellow-Red. 

Cu2 O + 2 {K 0, S O3) + Aq. 

Heated together. 

855. 2 (Cu 0, S O3) + 2 (Na O, C O2) -f C = 2 Cu 

+ 2 (Na O, S O3) + 3 C O2* 

856. Zn + Gu 0, S O3 -\- Aq = Cu ^ Zn 0, S O3 

+ Aq. 

357. Cu 0, H O + H = Cu + 2 H O. 

Black. Green Solution. 

359. CnO-^HOl-\-Aq= Cu Cl + Aq. 

Blue Solution. 

360. 3 Cu + 4 (^ (9, i\r Og) + «y = 3 (Cic 0, JSf 0,) 

+ a^ + TV O2. 



* The oxygen is removed by adding grape sugar to the solution. 



106 HEAVY METALS. 

Blue Salt. Black. White. 

360. 2(CuO, NO5. 4H0) -f Sn=2CuO + Sn02 

2 ((7m 0, S O3) -{- 2 {Fa 0, G 0,) -\- Aq = 

Blue or Green. 

Cu O, H O. Cu 0, C O2 + 2 {Na 0, S'O,) 

+ ^? + co,. 

3 (Cu 0, S 0,) -^IHO, 2 Na 0], POs + Aq== 

Greenish-Blue. 

3 Cu O, P O5 + 2 {Na 0, S 0,) + H 0, S 0, 

JrAq. 

Emerald-Green. 

Dioptase = 3 Cu 0, 2 Si O3. 3 H 0. 

361. Basic Acetate of Copper = 

Green. 

Cu 0, [C4 H3] O3. Cu O, H O. 5 H O. 

Neutral Acetate of Copper = 

Green. 

CuOCaHsiOg. 5 HO. 

Blue Solution. Black. 

362. GuO,SO^-^rHS-\-Aq=Cvi^-\-HO,SO^ 

-\-Aq. 

Black. Green Solution. 

CxxS^HGl-{-ciq= Ou Gl-i-aq-^JS.^. 

Mercury (Hg). 

367. Q Bff -{- A (ff 0, N Os) -i- aq = S (ffff^ 0, N Os) 

+ «^ + ]¥ O2. 

Black. 

368. Hff^ 0, NO, -{- K 0, H -\- Aq == Hg^ 

J^KO, N0, + Aq. 



HEAVY METALS. 107 

368. Hahnemann's Suboxide of Mercury has not a 
constant composition. Some chemists, however, 
assign to it the symbol 2 Hga 0, N O5. N H3. 

White. 

370. Hg^ 0, N 0^ -\- Na CI ^ Aq = Hg^ CI 

-\-Na 0, N 0,-[- Aq. 

371. d Hg-^4.{H0,N 0,)-\-aq = d{Hg 0,N 0,) 

+ aq-\-N O,. 

Yellow. 

ITg 0, NO, -{- KO, IT -{- Aq = Hg 

+ K O, N O5 + Aq. 

When heated. Red. 

372. Hg O, N O5 = Hg O + ]¥ O4 + O. 

373. ffg 0, N Os -{- Na Gl -\- Aq, gives no precipitate, 

because Hg CI is soluble in water. 

White. 

B.g O -\- IT CI -\- aq = Ug CI -{- IT -{- aq. 

Heated together. Residue. Sublimate. 

Hg O, S O3 + Na CI = Na O, S O3 + Hg CI. 

Black. 

Hg2 CI + ^ 0, ITO -\- Aq== Hg^O-^-K CI 

+ Aq. 

Yellow. 

Hg CI + .AT 0, i^ -i- ^^ = Hg O + ^ (7/ 

+ Aq. 

White. 

374. 2 Bg Gl-\- [iV^iT,] 0, HO-\-Aq = {n^^ X ci 

-\-Ha-\-Aq. 

Black Powder. 

375. Hg Cl-^ Sn Gl-\-Aq = Hg + Sn Ck + Aq. 

Black. 

376. HgCl + HS-^-Aq^-S-g^^ HCl-[-Aq. 



108 HEAVY METALS. 

Silver (Ag). 

380. 3 Ag + 4 {HO, NO,)-{-Aq = Z {Ag 0, NO,) 

+ ^^ + W O2. 

Heated together. 

381. a. Ag O, N O5 + 2 C = Ag + 2 C Oa + W O2. 

Brown, 

h. AgO,NO,-\-KO,irO+Aq = AgO,E.O 
-i-KO, NOs + Aq. 

White. 

Ago, nO-\-lNH,-\ 0, HO + Aq== {^Ag} 
+ Aq. 

White. 

c. Ag 0, N 0, -^ Na Gl -\- Aq ^ Ag CI 
+ iVa 6>, A^ O5 + Aq. 

Black 

e. AgO, NO,-\-HS-\-Aq^Ag^-\-HO, N 0, 

-{-Aq, 

Gold (Au). 

385. Axi-\-ZHGl + HO, NO^ ■{- Aq=Au Ok 
+ ^g + W O2. 

Dark-Brown, 

387. Au Ok + 6 {Fe 0, S O3) + .4^ = Au -f Fe^ Ok 

+ 2{Fe, O3, 3^03)+^^. 

388. Aurate of Potassa (K O, Au O3 + 4 H 0) is a 

compound of oxide of potassium and teroxide 
of gold, in which the last plays the part of an 
acid. 



HEAVY METALS. 109 

Platinum (Ft). 

391, BVi -^ ^HGl -\- 2{H0, N 0^) -{■ Aq = 

3 P< C4 + ^g + 2 W O2. 

Yellow. 

392. lNH,-\ CI -\-PtCk^-Aq= [N HJ CI, Pt Cl^ 

+ Aq. 

Yellow. 

394, K 01 + Pt Ok -\-Aq=K CI, Pt CI2 + Aq. 

Black. 

The Oxides of Platinum are Pt 0* and Pt Oj. 

Greenish-Brown. Reddish-Brown. 

The Chlorides of Platinum are Pt CI and Pt CI2. 

Black. Black. 

The Sulphides of Platinum are Pt S and Pt Sj. 

Black. 

Pt 01 + H S -^ Aq = -2tS -{- H 01 ^ Aq. 

Black, 

Pt (74 + 2 ir5+ ^^ = Pt Sa + 2 iTCZ + Aq. 



THIRD GROUP OF THE HEAVY METALS. 

Chromium (Cr). 
397. The symbol of chrome iron ore is Fe O, Crg O3 ; 
but almost invariably a portion of the Fe O is 
replaced by Mg 0, and a portion of the Crj O3 



Heated togrether in contact with air. 

2 (Fe 0, Cra O3) + 2 (K O, C 0^) + 7 O = 



by AI2 O3. 

Heated t 

(Fe 0, Cra 

Fe^ O3 + 2 (K O, 2 Cr O3). + 2 C O2 

* Only known in combination with water. 
10 



110 HEAVY METALS. 

397. The above process is hastened by mixing with the 
pulverized mineral a portion of nitre, which yields 
when heated a large supply of oxygen. 

Red. Yellow. 

898. K0,2Cr 0^-^K0, O0^^Aq=2 {KO, Cr O3) 
+ 4? + C O2. 

Yellow. Red. 

2 (KO, Cr 0,)-\-HO,NOs + Aq=KO, 2 Cr O3 
J^KO, N0,-\- Aq. 

399. K 0, Cr O3 + Pb 0, IC, H,'] 0^ -\- Aq =. 

Yellow. 

Pb O, Cr O3 + ^ 0, [C; ^3] O.^Aq. 

Yellow. Red. 

- 2 (PbO, Cr03) -^KO, HO-\-Aq = 2 Pb 0, Cr O3 
+ K0, Cr 6>3 + Aq. 

Yellow. White. 

400. 2 (Pb O, CvO,) -^ S H CI -\- Aq = 2 Pb CI 

Green. 

+ Cr^ Ck + .4^ + 3 CI. 

Cr^ Ck + 3 {IN H,-\ 0, H 0) + Aq = 
Cra O3, 10 H O + 3 [Nil,-] CI + Aq. 

KO, 2 Cr Os-\- H 0, S 0,-\- ^ S C,-\- Aq = 
KO, S 0,. Cr^ 6>3, 3 ^ O3 + ^^. 

The symbol of crystallized chrome alum is 
KO, SO3. Cr^Os, 3SO3. 24 HO. 

401. KO, 2 Cr Os -\- X* {JI 0, S0,)-\-aq=2 CrOg 

-\- KO, SO^. H 0, S Os + X {IT 0, S 0,) 
-\-aq. 

* X is here used to express an indefinite amount. 



HEAVY METALS. Ill 

Red. Green. 

401. a. 2 Cr O3 — 3 O = Cv^ O3. 

The oxygen in the last reaction may be removed by 
alcohol or any other reducing agent. 

Antimony (Sb). 
The oxides of antimony are as follows : — 

White. 

403. Oxide of Antimony, Sb O3. 

White. 

Antimonious Acid, Sb O4 = ^ (Sb O3, Sb O5). 

Pale-Yellow. 

Antimonic Acid, Sb O5. 

404. 3Sb+4(jy(9,i\^C»5)=3Sb04+4HO+4]\O2. 
When antimony is treated with an excess of con- 
centrated nitric acid, only Sb O4 appears to be 
formed. If, however, the acid is dilute, the anti- 
monious acid is mixed with more or less of basic 
nitrate of antimony (2 Sb O3, N O5) according to 
the degree of dilution. 

By heating together one part of metallic antimony 
and four parts of nitre in a crucible, there is 
formed a white mass, which is a mixture of anti- 
moniate of potassa (K 0, Sb O5) ; nitrite of 
potassa (K O, N O3) ; and undecomposed nitre 
(K O, N O5). Warm water will dissolve the 
two last, but not the antimoniate of potassa. If, 



112 HEAVY METALS. 

however, this anhydrous salt is boiled with water 
for one or two hours, it combines with five equiv- 
alents of water, forming a soluble compound 
(K O, Sb O5 . 5 H 0). The white mass, which 
seemed at first insoluble, dissolves in great meas- 
ure, leaving in suspension only a small amount of 
binantimoniate of potassa. 

405. ^hS3-{-3irCl-^aq=SbCls-{-aq-}-Sn^. 
Sb + 3 ^ CZ + (II 0, N 0,) -^aq^Sb Ck 

+ ag + ]V O2. 

Sb + X CI = ^5 (74 + X CI. 

Sh Ck^- Aq = ^h O^^ B H CI -\- Aq. 

The precipitate which is first formed on diluting a 
concentrated solution of Sb CI3 with water, always 
contains some chloride with the oxide ; but by con- 
tinued washing with water, or still better, with a 
weak solution of Na 0, C Og, the whole will be 
converted into oxide. 

Sh Gk + ^^ = Sb O5 + 5 IT CZ + Aq. 

406. Sb O3 4- [^ 0, K 0] a H, 0,, + Aq = 

IKO, ShO,-] G,H, 0,,-\-Aq. 
The symbol of crystallized tartar emetic is 
[K 0, Sb O3] Cs H4 Oio. 2 H O. 



HEAVY METALS. 113 

407. IK 0, Sh Os] OsH, 0,0 + S JiS-\-Aq = Sb'Sj 

[IT 0, K 0] G, H, 0:o + Aq. 
Kermes mineral is an amorphous modification of 

Sb Sg. 

Bri^ht-Ycllow. 

Sb Ol,-{-5 ffS-\-Aq= Sh^,-{-5IICl-{-Aq. 
Golden Sulphuret is a mixture of Sb S3 and Sb S5 . 

Melted together. 

408. Sb S3 -j- 3 Fe = Sb + 3 Fe S. 



Arsenic (As). 

412. As -j- 3 O = As O3 (arsenious acid). 

413. 2 As O3 + 3 C = 2 As + 3 C O2. 

414. As O3 -\- K 0, IT -\- Aq = K 0, As O3 -}- Aq. 
a. K 0, As 0, -\- 2 (Ou 0, S O3) -{- Aq = 

2 Cu OrAs O3 + ^ 0, SO,. BO, S 0,-{- Aq. 
■ h. The symbol of Schweinfurth green is 
Cu O, [Q H3] O3. 2 Cu O, As O3. 

415. As O3 + 2 {H 0, N 0,) -[- aq = As 0, -\- aq 

+ 2 W O4. 
The symbol of crystallized binarseniate of potassa 
is [2 H O, K O] As O5. 

Yellow. 

416. As 03-\-d H S-\-Aq = As^s+M- 

Yellow. 

As 0,-\-?> HS^ Aq= A&^,-\- Aq. 

Yellow. Red. 

2 As S3 + S = 2 As Sj. 
10* 



114 HEAVY METALS. 

416. The symbol of Mispickel (arsenical pyrites) is 

Fe [As, SJ. 
2 (Fe [As, S^]) + 13 O = Fea O3 + 2 As O3 
+ 2SO2. 

417. As Zng + 3 {HO, SO,)-\-Aq^Z {Zn 0, S 0,) 

+ ^9' + As H3. 

418. Sb Zng + 3 (^0, aS O3) + ^^ = 3 {Zk 0, S O3) 

^Aq-\-Sh Ha. 
The compounds of arsenic and antimony with hy- 
drogen are always mixed with more or less free 
hydrogen. When prepared as described in 
sections 417, 418 of Stockhardt's Elements, the 
gas consists almost entirely of hydrogen, con- 
taining only a very minute amount of either me- 
tallic compound. 



TABLES. 



EXPLANATION OF TABLES. 



Table I. — This table, Tvhich has been reprinted from the " Elementary 
Instructions in Chemical Analysis" by Fresenius, indicates by means of 
figures the solubility or insolubility in water and acids of some of the more fre- 
quently occurring compounds; thus, 1 means a substance soluble in water; 
2, a substance insoluble in water, but soluble in chlorohydric or nitric acid; 
8, a substance insoluble either in water or acids. For those substances stand- 
ing on the limits between these three classes, the figures are jointly expressed; 
thus 1-2 signifies a substance difficultly soluble in water, but soluble in 
chlorohydric or nitric acid ; 1 - 3, a body difficultly soluble in water, and the 
solubility of which is not increased on the addition of acids ; and 2 - 3, a sub- 
stance insoluble in water, and difficultly soluble in the acids. When the rela- 
tion of a substance to hydrochloric acid is different from that to nitric acid, 
this is stated in the notes. The figure indicating the solubility of a given salt 
will be found opposite to the symbol of its acid, in the column headed by the 
symbol of its base; that of a given binary, under the symbol of the correspond- 
ing oxide, and opposite to the symbol of its electro-negative element. 

Table II. — The values of the French measures and weights, in terms of the 
corresponding English units, given in this table, were taken from the second 
volume of the Cavendish Edition of " Gmelin's Hand-Book of Chemistry." 
The logarithms of these values and their arithmetical complements have been 
added to facilitate the reduction from one system to the other. The use of the 
table can be illustrated best by a few examples. 

1. It is required to reduce 560.367 metres to English feet. 
Solution. — No. of feet = No. of metres X No. of feet in one metre. 

log. No. of feet =: log. No. of metres + log. No. of feet in one metre, 
log. 560.367 2.7484726 

log. 3.2809 (value in feet of one metre from Table II.) 0.5159930 

" 3.2644656 

Ans. = 1838.51 feet. 



118 EXPLANATION OF TABLES. 

2. It is required to reduce 30.964 inches to centimetres. 

Solution. — No. of centimetres = No. of inches -^ No. of inches in one cen- 
timetre, 
log. No. of centimetres = log. No. of inches + log. (ar. co.) No. of 

inches in one centimetre, 
log. 30.964 1.4908571 

log. (ar. CO.) 0.3937 (value of one centimetre in inches) 0.4048^58 

1.8956829 
Ans. 78.6471 centimetres. 

3. It is required to reduce 23.576 kilometres to feet. 
Solution. — 23.576 kilometres = 23576 metres. 

No. of feet = No. of metres X No. of feet in one metre. ' 

log. 2.3576 4.3724701. 

log. 3.2809 0.5159930 

4.8884631 
Ans. 77350.5 feet. 

In the above examples logarithms of seven places have been used ; but where 
great accuracy is not required, logarithms of four places are sufficient. In 
such cases the last three figures of the logarithm given in the table may be 
neglected, and the problems solved with great expedition by means of the table 
of four-place logarithms which accompanies this book. 

Table III. — This table has been taken, with some few alterations, from 
Weber's " Atomgewichts-Tabellen." The atomic volumes assigned to the ele- 
ments are the same as those generally given in English and American text- 
books on Chemistrj', with the exception of those of Carbon, Boron, and Silicon, 
which are assumed to yield a one-volume gas like oxygen for convenience in 
calculation. The calculated specific gravities are deduced from the observed 
specific gravity of ox3'gen and the chemical equivalent of the given substance 
by means of the proportion, 

Equiv. of Oxygen : Equiv. of given substance = Sp. Gr. of Oxygen : Sp. Gr. 
of given substance. 

This proportion yields the specific gravity directly when one equivalent of 
the substance occupies the same volume as one equivalent of oxygen. If it 
occupies twice, three times, or four times this volume, the results must be 
divided by two, three, or four, as the case may be. The method of calculating 
may best be illustrated by a few examples. 

1. It is required to calculate the Specific Gravity of Nitrogen. ' 

Equiv. of 0. Equiv. ofN. Sp. Gr. of O. 

Solution. 8 : 14 = 1.10563 : 1.93485. 

This would be the specific gravity if 14 parts of nitrogen occupied the same 
volume as 8 parts of oxygen ; or, in other words, if the equivalent volume of 



EXPLANATION OF TABLES. 119 

nitrogen was 1, the same as that of oxygen. The fact is that it is 2, so that 
the true specific gravity of nitrogen = t (1.93485) = 0.967428. 

2. It is required to calculate the specific gravity of ammonia gas. 

EquiT. ofO. Equiv. ofNHj. Sp. Gr. ofO. 

Solution. 8 : 17 = 1.10563 : 2.34946. 

Hence the specific gravity of ammonia gas •would be 2.34946 if one equiva- 
lent (or 17 parts) occupied the same volume as one equivalent of oxygen; but 
on referring to the table, it will be found that the equivalent volume of this gas 
is 4, or, in other words, that one equivalent occupies a volume four times as 
large as that occupied by one equivalent of oxygen, so that to find the true 
specific gravity of ammonia gas we must divide 2.34946 by 4, which will give 
a quotient 0.58736. The slight difference between this result and that given 
in the table arises from the fact that in Weber's table the equivalent of nitro- 
gen used is 14.005, and not 14, as in the solutions of the above examples. 

Were the law of equivalent volumes absolutely rigorous, (that is, did one 
equivalent of every gas precisely occupy either the same volume, or else a vol- 
ume two, three, or four times as great as the volume of oxygen,) then the cal- 
culated specific gravities ought to agree exactly with those obtained by experi- 
ment. On comparing together the two columns of observed and calculated 
specific gravities in the table, it will be found that the numbers, although 
approximatively equal, do not absolutely coincide. Part of these differences 
are unquestionably owing to errors of observation ; but after making the great- 
est possible allowance for all errors of that sort, there still remains (especially 
in the case of the easily condensed gases, such as alcohol vapor, sulphurous 
acid, and carbonic acid) large differences to be accounted for. The most prob- 
able explanation of these differences seems to be found in the assumption that 
the law of equivalent volumes holds rigorously only when the gases are in the 
state of extreme expansion. As we experiment upon them, they are more or 
less condensed by the pressure of the atmosphere, and it is supposed that they 
are not all condensed equally, or, in other words, that even under this pressure 
they do not obey absolutely the law of Marriotte. The more easily a gas may 
be reduced to a fluid, the greater is it condensed by the atmospheric pressure, 
and hence the greater is its specific gravity. This view is confirmed by the 
fact that the observed specific gravity of carbonic acid gas at 0°, and under a 
more feeble pressure than that of the atmosphere, approaches more nearly to 
that obtained by experiment. 

The Specific gravity of Carbonic Acid Gas, at 0° (air = 1), was. 

Under the pressure of 76.000 centimetres, 1.52910 

" " " 37.413 " 1.52366 

" " " 22.417* " 1.52145 

Theoretical specific gravity, 1.52024 

* It must be remembered that the specific gravity of a gas is equal to its weight, di- 
vided by the weight of an equal volume of air under the same conditions of temperature 
and pressure. 



120 



TABLES. 



^ I 

O 
I— I 

iJ 
O 



o 

i-Ih 


CO G<J (N (N 
(M |<MCO |<>?THffq<N(>»(MTHG<l | | 

CN 1—1 tH I— 1 


o 


l-H 


o 

s 




O 


1—1 t-l 


o 


(MrHrH(MT-( r-l(MCCi G<><J^G<JfH(M 


O 


<Mi-(r-l(Mi-lrHT-l(M(M (M(N(NtH | 


o 


CO CO 
r-l • CM 


O 


tHrHT-lTHC0<?^TH<M<N<N5<l<?^(n rH <M 


o 

in 


tHrHTHrHCO(Mi-l<MCNi-H(NCN(M i-l (M 


o 


(M (M CO 

T-l iH 1— 1 


o 

be 

<1 


jcoco<N |g^ih<m(N(»(M(Mit^ ih cm 

r-t r-l 


o 
M ■ 




o 




o 

1 — 1 
1—1 






OhHC/2ccO^ptH<|<lOpq0|<1lH 



121 





(M T-i 


(MrH<>?iH (M (N(?^C<Ir-((M 




++ 


<M CN C<1 (Ja <M 1 i-H i-H 


6 


tH 




O 


5<l i-H 


CO l-H t-^ 


6 


CO 


++ 


O 


|>1 tH 


<>? tH <M (M <3<1 iH 1 


6 


CO 


1— lr-liHG<l(M (m|i— liH 
tH 


O 

J? 


CO 


!H(NtH<Mi-l G<Jr-lrHTH 


c5 


CO 


iHCNrHCMfKI «<l(?qr-<r-l 


o 


(M j-l 


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o 


(N 1—1 




o 


tH r-l 


o 

1 


<Mt-(THS<JtHtHiH(M(M (I<1(M<M | 




Om 









is 




















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:S 


















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y. 


o 










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03 


pi 










t>i 








rO 








O 


ro 








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a 


d 
















rS 




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43 






'« 


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rC 


o 


03 






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CJ 


^rO 


a 


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o 

■73 


• l-H 


1^ 


>% 


r" 


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C) 




n 


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s 


s 


t> 




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rQrQ 


13 


o 


^i 


'C itJ" 




o 


-H 


•ri 


•p! 


^ 


'« 


— 1 






~ 


(=-.o 


o 


<■> 


a 

n1 


.£3 


s 


*r^ 




h 




'S 


'S 


13 




d 
o 


d 


fl 


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<D 


o 


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o 


n 






-> 


ccco 


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11 



122 



TABLES. 



TABLE II. 



FKENCH MEASUEES AND WEIGHTS. 
Measures of Length. 



1 Kilometre 

1 Hectometre 

1 Decimetre 

1 Metre 



1 Kilometre 

1 Metre 

1 Centimetre 



1000 Metres. 


100 « 


10 " 


1 " 


0.6214 Miles. 


3.2809 Feet. 


0.3937 Inches 



1 Meti-e = 1.000 Metre. 

1 Decimetre = 0.100 " 

1 Centimetre = 0.010 " 

1 MilUmetre = 0.001 " 



Logarithms. 
9.7933712 
0.5159930 
9.5951742 



Ar. Co. Log. 
0.2066188 
9.4840070 
0.4048258 



Measures of Volume. 

1 Cubic Metre = 1000.000 Litres. 

1 Cubic Decimetre = 1.000 " 

1 Cubic Centimetre = 0.001 " 







Logarithms. 


Ar. Co. Log, 


1 Cubic Metre 


= 35.31660 Cubic Feet. 


1.5479790 


8.4520210 


1 Cubic Decimetre 


= 61.02709 Cubic Inches. 


1.7855226 


8.2144774 


1 Cubic Centimetre 


= 0.06103 " 


8.7855226 


1.2144774 


1 Litre 


= 0.22017 Gallons. 


9,3427581 


0.6572419 


1 Litre 


= 0.88066 Quarts. 


9.9448083 


0.0551917 


1 Litre 


= 1.76133 Pints. 
Weights. 


0.2458407 


9.7541593 



1 Kilogramme = 1000 Grammes. 

1 Hectogramme = 100 " 

1 Decagramme =10 " 

1 Gramme = 1 " 



1 Gramme = 1,000 Gramme. 

1 Decigramme = 0.100 " 

1 Centigramme = 0.010 " 

1 Milligramme = 0.001 " 



1 Kilogramme = 2.67951 Pounds (Troy). 
1 Gramme = 15.44242 Grains. 



Logarithms. Ar, Co. Log, 
0.4280554 9.5719446 
1.1887154 8.8112846 



TABLES. 



123 



TABLE III. 

SPECIFIC GRAVITY AND ABSOLUTE WEIGHT OF ONE 
LITRE OF SOME OF THE MOST IMPORTANT GASES 
AND VAPORS. 



Names of Gases. 


> 
'3 


Sp. Gr. 
Observed. 


Sp. Gr. 
Calculated. 


Weight of 
1 Litre = 
1000 c. c. 


Loga- 
rithms. 


Ar. Co. Log- 
arithms. 


Air, 




1. 


1.00000 


1.29363 


0.1118101 


9.8881899 


Alcohol, 


4 


1.613 


1.58934 


2.05602 


0,3130273 


9.6869727 


Ammonia Gas, 


4 


0.5967 


0.58753 


0.76005 


9.8808422 


0.1191578 


Antimony, 


1 




17.83274 


23.06897 


1.3630282 


8.6369718 


Antimonide of Hydr. 


4 




4.56239 


5.90204 


0.7710022 


9.2289978 


Arsenic, 


1 


10.65 


10.36528 


13.40884 


1.1273912 


8.8726088 


Arsenide of Hydr. 


4 


2.695 


2.69553 


3.48702 


0.5424519 


9.4575481 


Boron, 


1 




1.50591 


1.94809 


0.2896090 


9.7103910 


Bromine, 


2 


5.54 


5.52605 


7.14866 


0.8542247 


9.1457753 


Bromoliydric Acid, 


4 




2.79758 


3.61903 


0.5585922 


9.4414078 


Carbon, 


1 


0.8469* 


0.82922 


1.07270 


0.0304783 


9.9695217 


Carbonic Acid, 


2 


1.52908 


1.52024 


1.96663 


0.2937226 


9.7062774 


Carbonic Oxide, 


2 


0.96779 


0.96743 


1.25150 


0.0974309 


9.9025691 


Chlorine, 


2 


2.47 


2.45052 


3.17007 


0.5010689 


9.4989311 


Chloride of Boron, 


4 


3.942 


4.05226 


5.24213 


0.7195078 


9.2804922 


Chloride of Silicon, 


3 


5.939 


5.92477 


7.66446 


0.8844816 


9.1155184 


Chlorohydric Acid, 


4 


1.2474 


1.25981 


1.62973 


0.2121157 


9.7878843 


Cyanogen, 


2 


1.8064 


1.79698 


2.32463 


0.3663538 


9.6336462 


Cyanhydric Acid, 


4 


0.9476 


0.93304 


1.20701 


0.0817109 


9.9182891 


Ether, 


2 


2.586 


2.55677 


3.30751 


0.5195012 


9.4804988 


Fluorine, 


2 




1.30151 


1.68367 


0.2262570 


9.7737430 


Fluoride of Boron, 


4 


2.3124 


2.32875 


3.01254 


0.4789329 


9.5210671 


Fluoride of Silicon, 


3 


3.600 


3.62677 


4.69170 


0.6713302 


9.3286698 


Fluohydric Acid, 


4 




0.68531 


0.88654 


9.9476983 


0.0523017 


Hydrogen, 


2 


0,06927 


0.06910 


0.08939 


8.9512889 


1.0487111 


Iodine, 


2 


8.716 


8.76760 


11.34203 


1.0547011 


8.9452989 


lodohydric Acid, 


4 


4.443 


4.41835 


5.71571 


0.7570702 


9.2429298 


Marsh Gas, 


4 


0.5576 


0.55282 


0.71514 


9.8543911 


0.1456089 


Mercuiy, 


2 


6.976 


6.91732 


8.94845 


0.9517478 


9.0482522 


Nitrogen, 


2 


0.97136 


0.96776 


1.25192 


0.0975765 


9.9024235 


Nitrous Oxide, 


2 


1.5269 


1.58951 


2.05624 


0.3130738 


9.6869262 


Nitric Oxide, 


4 


1.0388 


1.03669 


1.34109 


0.1274580 


9.8725420 


defiant Gas, 


4 


0.9852 


0.96743 


1.25150 


0.0974309 


9.9025691 


Oxygen, 


1 


l.li)§63 




1.43028 


0.1554210 


9.8445790 


Phosphoms, 


1 


4.42 


4.33452 


5.60727 


0.7487515 


9.2512485 


Phosphide of Hydr. 


4 


1.178 


1.18728 


1.53590 


0.1863629 


9.8136371 


Selenium, 


1 




2.73801 


3.54197 


0.5492448 


9.4507552 


Silicon, 


1 




3.07120 


3.97300 


0.5991186 


9.4008814 


Sulphur, 


k 


6.5635 


6.65866 


8.61384 


0.9351968 


9.064S032 


Sulphide of Hydr. 


2 


1.1912 


1.17888 


1.52.503 


0.1832784 


9.8167216 


Sulphurous Acid, 


2 


2.247 


2.21541 


2.86592 


0.4572640 


9.5427360 



* Calculated from the Sp. Gr. of Carbouic Acid Gas, observed by RegnauU. 



124 



TABLES. 



TABLE IV. 

MEAN COEFFICIENTS OF LINEAE EXPANSION OF SOLIDS 
FOR ONE DEGEEE BETWEEN 0° AND 100°. 



Name of Substance. 


Coefficients. 


Name of Observer. 


Glass (flint of Clioisy le Roi), 


0.00000760 


Regnault. 


Platinum, 


0.00000884 


Dulong and Petit. 


Glass (commoa of Paris), 


0.00000920 


Regnault. 


Palladium, 


0.00001000 


WollastOn. 


Antimony, 


0.00001083 


Smeaton. 


Iron (soft forged), 


0.00001220 


Lavoisier and Laplace. 


Bismuth, 


0.00001392 


Smeaton. 


Gold, 


0.00001466 


Lavoisier and Laplace. 


Brass (English, in rods), 


0.00001893 


Roy. 


Copper, 


0.00001919 


Troughton. 


Silver, 


0.00002083 


« 


Tin (fine), 


0.00002283 


<i 


Lead, 


0.00002866 


a 


Zinc, 


0.00002942 


a 



APPARENT CUBIC EXPANSION OF LIQUIDS IN GLASS 
BETWEEN 0° AND 100°. 



Name of Substance. 


Expansion from Oo to lOOo. 


Water, 


A 


— 


0.0466 


Chlorohydric Acid, Sp. Gr. 1.137, 


2V 


= 


0.0600 


Nitric Acid, Sp. Gr. 1.40, 


* 


= 


0.0100 


Sulphuric Acid, Sp. Gr. 1.85, 


tV 


= 


0.0600 


Common Ether, 


-h 


= 


0.0700 


Olive Oil, 


-h 


= 


0.0800 


Oil of Tui-pentine, 


-h 


= 


0.0700 


Water saturated with salt, 


■h 


= 


0.0500 


Alcohol, 


1 
9 


= 


0.1100 


Mercury, 


-h 


= 


0.0156 



COEFFICIENTS OF CUBIC EXPANSION OF GASES. 
Observed bt V. REGNAULT. 



Name of Substance. 


Under constant Volume. 


Under constant Pressure. 


Air, 


0.003665 


0.003670 


Nitrogen, 


0.003668 


0.003670 


Hydrogen, 


0.003667 


0.003661 


Oxide of Carbon, 


0.003667 


0.003669 


Carbonic Acid, 


0.003688 


0.003710 


Cyanogen, 


0.003829 


0.003877 


Sulphm-ous Acid, 


0.003845 


0.003903 



TABLES. 



125 



TABLE V. 

DENSITIES AND VOLUMES OF WATER. 

By M. DESPRETZ. 



as 

S3 


Volumes. 


Densities. 




Volumes. 


Densities. 




Volumes. 


Densities. 





1.0001269 


0.999873 


34 


1.00555 


0.994480 


68 


1.02144 


0.979010 


1 


1.0000730 


0.999927 


35 


1.00593 


0.994104 


69 


1.02200 


0.978473 


2 


1.0000331 


0.9999661 


36 


1.00624 


0.993799 


70 


1.02255 


0.977947 


3 


1.0000083 


0.999999 


37 


1.00661 


0.993433 


71 


1.02315 


0.977373 


4 


1.0000000 


1.000000 


38 


1.00699 


0.993058 


72 


1.02375 


0.976800 


5 


1.0000082 


0.999999 


39 


1.00734 


0.992713 


73 


1.02440 


0.976181 


6 


1.0000309 


0.999969 


40 


1.00773 


0.992329 


74 


1.02499 


0.975619 


7 


1.0000708 


0.999929 


41 


1.00812 


0.991945 


75 


1.02562 


0.975018 


8 


1.0001216 


0.999878 


42 


1.00853 


0.991542 


76 


1.02631 


0.974364 


9 


1.0001879 


0.999812 


43 


1.00894 


0.991139 


77 


1.02694 


0.973766 


10 


1.0002684 


0.999731 


44 


1.00938 


0.990707 


78 


1.02761 


0.973132 


11 


1.0003598 


0.999640 


45 


1.00985 


0.990246 


79 


1.02823 


0.972545 


12 


1.0004724 


0.999527 


46 


1.01020 


0.989903 


80 


1.02885 


0.971959 


13 


1.0005862 


0.999414 


47 


1.01067 


0.989442 


81 


1.02954 


0.971307 


14 


1.0007146 


0.999285 


48 


1.01109 


0.989032 


82 


1.03022 


0.970666 


15 


1.0008751 


0.999125 


49 


1.01157 


0.988562 


83 


1.03090 


0.970027 


16 


1.0010215 


0.998979 


50 


1.01205 


0.988093 


84 


1.03156 


0.969405 


17 


1.0012067 


0.998794 


51 


1.01248 


0.987674 


85 


1.03225 


0.968757 


18 


1.00139 


0.998612 


52 


1.01297 


0.987196 


86 


1.03293 


0.968120 


19 


1.00158 


0.998422 


53 


1.01345 


0.986728 


87 


1.03351 


0.967482 


20 


1.00179 


0.998213 


154 


1.01395 


0.986243 


88 


1.03430 


0.966837 


21 


1.00200 


0.998004 


55 


1.01445 


0.985756 


1 89 


1.03500 


0.966183 


22 


1.00222 


0.997784 


56 


1.01495 


0.985270 


90 


1.03566 


0.965567 


23 


1.00244 


0.997566 


57 


1.01547 


0.984766 


91 


1.03639 


0.964887 


24 


1.00271 


0.997297 


58 


1.01597 


0.984281 


92 


1.03710 


0.964227 


25 


1.00293 


0.997078 


59 


1.01647 


0.983798 


93 


1-03782 


0.963558 


26 


1.00321 


0.996800 


60 


1.01698 


0.983303 


94 


1.03852 


0.962908 


27 


1.00345 


0.996562 


61 


1.01752 


0.982782 


95 


1.03925 


0.962232 


28 


1.00374 


0.996274 


62 


1.01809 


0.982231 


96 


1.03999 


0.961547 


29 


1.00403 


0.995986 


63 


1.01862 


0.981720 


97 


1.04077 


0.960827 


30 


1.00433 


0.995688 


64 


1.01913 


0.981229 


j 98 


1.04153 


0.960125 


31 


1.00463 


0.995391 


65 


1.01967 


0.980709 


99 


1.04228 


0.959434 


32 


1.00494 


0.995084 


66 


1.02025 


0.980152 


jlOO 


1.04315 


0.958634 


33 1 1.00525 


0.994777 


67 


1.02085 


0.979576 









126 



TABLES. 



TABLE VI. 

PER CENT OF N O5 IN AQUEOUS SOLUTIONS OF DIF- 
FERENT SPECIFIC GRAVITIES. By URE. 15° C. 



Specific 


Per Cent 


Specific 


Per Cent 


Specific 


Per Cent 


Specific 


Per Cent 


Gravity. 


of N O5. 


Gravity. 


ofNOg. 


Gravity. 


of N Og. 


Gravity. 


ofNOj. 


1.500 


79.7 


1.419 


59.8 


1.295 


39.8 


1.140 


19.9 


1.498 


78.9 


1.415 


59.0 


1.289 


39.0 


1.134 


19.1 


1.496 


78.1 


1.411 


58.2 


1.283 


38.3 


1.129 


18.3 


1.494 


77.3 


1.406 


57.4 


1.276 


37.5 


1.123 


17.5 


1.491 


76.5 


1.402 


56.6 


1.270 


36.7 


1.117 


16.7 


1.488 


75.7 


1.398 


55.8 


1.264 


35.9 


1.111 


15.9 


1.485 


74.9 


1.394 


55.0 


1.258 


35.1 


1.105 


1.5.1 


1.482 


74.1 


1.388 


54.2 


1.252 


34.3 


1.099 


14.3 


1.479 


73.3 


1.383 


53.4 


1.246 


33.5 


1.093 


13.5 


1.476 


72.5 


1.378 


52.6 


1.240 


32.7 


1.088 


12.7 


1.473 


71.7 


1.373 


51.8 


1.234 


31.9 


1.082 


11.9 


1.470 


70.9 


1.368 


51.1 


1.228 


31.1 


1.076 


11.2 


1.467 


70.1 


1.363 


50.2 


1.221 


30.3 


1.071 


10.4 


1.464 


69.3 


1.358 


49.4 


1.215 


29.5 


1.065 


9.6 


1.460 


68.5 


1.3.53 


48.6 


1.208 


28.7 


1.059 


8.8 


1.457 


67.7 


1.348 


47.9 


1.202 


27.9 


1.054 


8.0 


1.453 


66.9 


1.343 


47.0 


1.196 


27.1 


1.048 


7.2 


1.450 


66.1 


1.338 


46.2 


1.189 


26.3 


1.043 


6.4 


1.446 


65.3 


1.332 


45.4 


1.183 


25.5 


1.037 


5.6 


1.442 


64.5 


1.327 


44.6 


1.177 


24.7 


1.032 


4.8 


1.439 


63.8 


1.322 


43.8 


1.171 


23.9 


1.027 


4.0 


1.435 


63.0 


1.316 


43.0 


1.165 


23.1 


1.021 


3.2 


1.431 


62.2 


1.311 


42.2 


1.159 


22.3 


1.016 


2.4 


1.427 


61.4 


1.306 


41.4 


1.153 


21.5 


1.011 


1.6 


1.423 


60.6 


1.300 


40.4 


1.146 


20.7 


1.005 


0.8 



PER CENT OF H CI IN AQUEOUS SOLUTIONS OF DIF- 
FERENT SPECIFIC GRAVITIES. By E DAVY 15° C. 



Specific Gravity. 


Per Cent of H CI. 


1 

i SiJecific Gravity, 


Per Cent of H CI. 


1.21 


42.43 


110 


20.20 


1.20 


40.80 


1.09 


18.18 


1.19 


38.38 


1.08 


16.16 


1.18 


36.36 


1.07 


14.14 


1.17 


34.34 


1.06 


12.12 


1.16 


32.32 


1.05 


10.10 


1.15 


30.30 


1.04 


8.08 


1.14 


28.28 


1.03 


6.06 


1.13 


26.26 


1.02 


4.04 


1.12 


24.24 


1.01 


2.02 


1.11 


22.22 







TABLES. 



127 



PER CENT OF HO, S O3 AND S O3 IN AQUEOUS SO- 
LUTIONS or DIEEEEENT SPECIFIC GEAVITIES. By 
BINEAU. 15° C. 



Per Cent of 


Specific 


Per Cent of 


Percent of 


Specific 


Per Cent of 


HO, SO3. 


Gravity. 


SO3. 


H 0, S O3 


Gravity. 


SO3. 


100 


1.8426 


81.63 


56 


1.4586 


45.71 


99 


1.842 


80.81 


55 


1.448 


44.89 


98 


1.8406 


80.00 


54 


1.438 


44.07 


97 


1.840 


79.18 


53 


1.428 


43.26 


96 


1.8384 


78.36 


52 


1.418 


42.45 


95 


1.8376 


77.55 


51 


1.408 


41.63 


94 


1.8356 


76.73 


50 


1.398 


40.81 


93 


1.834 


75.91 


49 


1.3886 


40.00 


92 


1.831 


75.10 


48 


1.379 


39.18 


91 


1.827 


74.28 


47 


1.370 


38.36 


90 


1.822 


73.47 


46 


1.361 


37.55 


89 


1.816 


72.65 


45 


1.351 


36.73 


88 


1.809 


71.83 


44 


1.342 


35.82 


87 


1.802 


71.02 


43 


1.333 


35.10 


86 


1.794 


70.10 


42 


1.324 


34.28 


85 


1.786 


69.38 


41 


1.315 


33.47 


i 84 


1.777 


68.57 


40 


1.306 


32.65 


! 83 


1.767 


67.75 


39 


1.2976 


31.83 


82 


1.756 


66.94 


38 


1.289 


31.02 


81 


1.745 


66.12 


37 


1.281 


30.20 


80 


1.734 


65.30 


36 


1.272 


29.38 


79 


1.722 


64.48 


35 


1.264 


28.57 


78 


1.710 


63.67 


34 


1.256 


27.75 


77 


1.698 


62.85 


33 


1.2476 


26.94 


76 


1.686 


62.04 


32 


1.239 


26.12 


75 


1.675 


61.22 


31 


1.231 


25.30 


74 


1.663 


60.40 


30 


1.223 


25.49 


73 


1.651 


59.59 


29 


1.215 


23.67 


72 


1.639 


58.77 


28 


1.2066 


22.85 


71 


1.637 


57.95 


27 


1.198 


22.03 


70 


1.615 


57.14 


26 


1.190 


21.22 


69 


1.604 


56.32 


25 


1.182 


20.40 


68 


1.592 


55.59 


24 


1.174 


19.58 


67 


1.580 


54.69 


22 


1.159 


17.95 


66 


1.578 


53.87 


20 


1.144 


16.32 


65 


1.557 


53.05 


18 


1.129 


14.69 


64 


1.545 


52.24 


16 


1.1136 


13.06 


63 


1.534 


51.42 


14 


1.098 


11.42 


62 


1.523 


50.61 


12 


1.083 


9.79 


61 


1.512 


49.79 


10 


1.068 


8.16 


60 


1.501 


48.98 


8 


1.0536 


6.53 


59 


1.490 


48.16 


6 


1.039 


4.89 


58 


1.480 


47.34 


4 


1.0256 


3.26 


57 


1469 


46.53 


2 


1.013 


1.63 



128 



TABLES. 



PER CENT OF NH3 IN AQUEOUS SOLUTIONS OF DIF- 
FERENT SPECIFIC GRAVITIES. By J. OTTO. 16° C 



Specific 


Per Cent of 


Specific 


Per Cent of 


Specific 


Per Cent of 


Gravity. 


NH3. 


Gravity. 


NH3. 


Gravity. 


NH3. 


0.9517 


12.000 


0.9607 


9.625 


0.9697 


7.250 


0.9521 


11.875 


0.9612 


9.500 


0.9702 


7.125 


0.9526 


11.750 


0.9616 


9.375 


0.9707 


7.000 


0.9531 


11.625 


0.9621 


9.250 


0.9711. 


6.875 


0.9536 


11.500 


0.9626 


9.125 


0.9716 


6.750 


0.9540 


11.375 


0.9631 


9.000 


0.9721 


6.625 


0.9545 


11.250 


0.9636 


8.875 


0.9726 


6.500 


0.9550 


11.125 


0.9641 


8.750 


0.9730 


6.375 


0.9555 


11.000 


0.9645 


8.625 


0.9735 


6.250 


0.9556 


10.950 


0.9650 


8.500 


0.9740 


6.125 


0.9559 


10.875 


0.9654 


8.375 


0.9745 


6.000 


0.9564 


10.750 


0.9659 


8.250 


0.9749 


5.875 


0.9569 


10.625 


0.9664 


8.125 


0.9754 


5.750 


0.9574 


10.500 


0.9669 


8.000 


0.9759 


5.625 


0.9578 


10.375 


0.9673 


7.875 


0.9764 


5.500 


0.9583 


10.250 


0.9678 


7.750 


0.9768 


5.375 


0.9588 


10.125 


0.9683 


7.625 


0.9773 


5.250 


0.9593 


10.000 


0.9688 


7.500 


0.9778 


5.125 


0.9597 


9.875 


0.9692 


7.375 


0.9783 


5.000 


0.9602 


9.750 











TABLE VII. 

PER CENT OF OXYGEN IN DIFFERENT OXIDES. 



Oxides. 


Per Cent. 


Logarithms. 


Oxides. 


Per Cent. 


Logaritlims. 


AI2O3 


0.467 


9.6684 


KO 


0.170 


9.2304 


AsOs 


0.348 


9.5416 


LiO 


0.550 


9.7404 


BO3 


0.688 


9.8376 


MO3 


0.343 


9.5353 


BaO 


0.104 


9.0170 


MgO 


0.400 


9.6021 


BeaOs 


0.630 


9.7993 


MnO 


0.225 


9.3522 


BiaOs 


0.103 


9.0128 


Mna O3 


0.303 


9.4814 


COa 


0.727 


9.8615 


NO5 


0.740 


9.8692 


Ca 


0.286 


9.4564 


NaO 


0.258 


9.4116 


CoO 


0.213 


9.3284 


NiO 


0.213 


9.3284 


Crs O3 


0.473 


9.6749 


PO5 


0.563 


9.7505 


CuaO 


0.112 


9.0492 


PbO 


0.717 


9.8555 


CuO 


0.201 


9.3032 


SiOs 


0.530 


9.7243 


FeO 


0.222 


9.3464 


SrO 


0.154 


9.1875 


FeaOs 


0.300 


9.4771 


TaOs 


0.115 


9.0607 


HO 


0.889 


9.9489 


ZnO 


0.197 


9.2945 



ANTILOGARITHMS. 


. a 






















Proportional Parts. | 


n 





1 


2 


3 


4 


5 


6 


7 


8 


9 










II 


































1019 




1 2 




s 


4 


5 


6 

1 


7 
2 


8 

2 


9 
2 


.00 


1000 


1002 


1005 


1007 


1009 


1012 


1014 


1016 


1021 






.01 


1023 


1026 1028:1030 


1033 


1035 


1038 1040 


1042 


1045 











1 


2 


2 


2 


.02 


1047 


1050 


1052 1054 


1057 


1059 


1062 


1064 


1067 


1069 











1 


2 


2 


2 


.03 


1072 


1074 


1076 


1079 


1081 


1084 


1086 


1089 


1091 


1094 











1 


2 


2 


2 


.04 


1096 


1099 


1102 


1104 


1107 


1109 


1112 


1114 


1117 


1119 


1 








2 


2 


2 


2 


.05 


1122 


1125 


1127 


1130 


1132 


1135 


1138 


1140 


1143 


1146 


1 








2 


2 


2 


2 


.06 


1148 


1151 


1153 


1156 


1159 


1161 


1164 


1167 


1169 


1172 


1 








2 


2 


2 


2 


.07 


1175 


1178 


1180 


1183 


1186 


1189 


1191 


1194 


1197 


1199 


1 








2 


2 


2 


2 


.08 


1202 


1205 


1208 


1211 


1213 


1216 


1219 


1222 


1225 


1227 


1 








2 


2 


2 


3 


.09 
.10 


1230 
1259 


1233 
1262 


1236 
1265 


1239 
1268 


1242 
1271 


1245 
1274 


1247 
1276 


1250 
1279 


1253 
1282 


1256 
1285 


1 
1 






1 


2 
2 


2 
2 


2 


3 


2 


3 


.11 


1288 


1291 


1294 


1297 


1300 


1303 


1306 


1309 


1312 


1315 


1 






2 


2 


2 


2 


3 


.12 


1318 


1321 


1324 


1327 


1330 


1334 


1337 


1340 


1343 


1346 


1 






2 


2 


2 


2 


3 


.13 


1349 


1352 


1355 


1358 


1361 


1365 


1368 


1371 


1374 


1377 


1 






2 


2 


2 


3 


3 


.14 


1380 


1384 


1387 


1390 


1393 


1396 


1400 


1403 


1406 


1409 


1 






2 


2 


2 


3 


3 


.15 


1413 


1416 


1419 


1422 


1426 


1429 


1432 


1435 


1439 


1442 


1 






2 


2 


2 


3 


3 


.16 


1445 


1449 


1452 


1455 


1459 


1462 


1466 


1469 


1472 


1476 


1 






2 


2 


2 


3 


3 


.17 


1479 


1483 


1486 


1489 


1493 


1496 


1500 


1503 


1507 1 1510 


1 






2 


2 


2 


3 


3 


.18 


1514 


1517 


1521 


1524 


1528 


1531 


1535 


1538 


1542' 1545 


1 






2 


2 


2 


3 


3 


.19 


1549 


1552 


1556 


1560 


1563 


1567 


1570 


1574 


1578 1581 


1 






2 


2 


3 


3 


3 


.20 


1585 


1589 


1592 


1596 


1600 


1603 


1607 


1611 


1614 1618 


1 




1 


2 


2 


3 


3 


3 


.21 


1622 


1626 


1629 


1633 


1637 


1641 


1644 


1648 


1652 1656 


1 




2 


2 


2 


3 


3 


3 


.22 


1660 


1663 


1667 


1671 


1675 


1679 


1683 


1687 


1690! 1694 


1 




2 


2 


2 


3 


3 


3 


.23 


1698 


1702 


1706 


1710 


1714 


1718 


1722 


1726 


1730 1734 


1 




2 


2 


2 


3 


3 


4 


.24 


1738 


1742 


1746 


1750 


1754 


1758 


1762 


1766 


1770 1774 


1 




2 


2 


2 


3 


3 


4 


.25 


1778 


1782 


1786 


1791 


1795 


1799 


1803 


1807 


1811 1816 


1 




2 


2 


2 


3 


3 


4 


.26 


1820 


1824 


1828 


1832 


1837 


1841 


1845 


1849 


1854 1858 


1 




2 


2 


3 


3 


3 


4 


.27 


1862 


1866 


1871 


1875 


1879 


1884 


1888 


1892 


1897 1901 


1 




2 


2 


3 


3 


3 


4 


.28 


1905 


1910 


1914 


1919 


1923 


1928 


1932 


1936 


1941 1 1945 


1 




2 


2 


3 


3 


4 


4 


.29 


1950 


1954 


1959 


1963 


1968 


1972 


1977 


1982 


1986; 1991 


1 




2 


2 


3 


3 


4 


4 


.30 


1995 


2000 


2004 


2009 


2014 


2018 


2023 


2028 


2032 


2037 


1 




2 


2 


3 


3 


4 


4 


.31 


2042 


2046 


2051 


2056 


2061 


2065 


2070 


2075 


2080 


2084 


1 




2 


2 


3 


3 


4 


4 


•32 


2089 


2094 


2099 


2104 


2109 


2113 


2118 


2123 


2128 


2133 


1 




2 


2 


3 


3 


4 


4 


.33 


2138 


2143 


2148 


2153 


2158 


2163 


2168 


2173 


2178 


2183 


1 




2 


2 


3 


3 


4 


4 


.34 


2188 


2193 


2198 


2203 


2208 


2213 


2218 


2223 


2228 


2234 


1 1 


2 


2 


3 


3 


4 


4 


5 


.35 


2239 


2244 


2249 


2254 


2259 


2265 


2270 


2275 


2280 


2286 


1 1 


2 


2 


3 


3 


4 


4 


5 


.36 


2291 


2296 


2301 


2307 


2312 


2317 


2323 


2328 


2333 


2339 


1 1 


2 


2 


3 


3 


4 


4 


5 


.37 


2344 


2350 


2355 


2360 


2366 


2371 


2377 


2382 


2388 


2393 


1 1 


2 


2 


3 


3 


4 


4 


5 


.38 


2399 


2404 


2410 


2415 


2421 


2427 


2432 


2438 


2443 


2449 


1 1 


2 


2 


3 


3 


4 


4 


5 


M 


2455 


2460 


2466 


2472 


2477 


2483 


2489 


2495 


2500 


2506 


1 1 


2 


2 


3 


3 


4 


5 


5 


.40 


2512 


2518 


2523 


2529 


2535 


2541 


2547 


2553 


2559 


2564 


1 1 


2 


2 


3 


4 


4 


5 


5 


.41 


2570 


2576 


2582 


2588 


2594 


2600 


2606 


2612 


2618 


2624 


1 I 


2 


2 


3 


4 


4 


5 


5 


.42 


2630 


2636 


2642 


2649 


2655 


2661 


2667 


2673 


2679 


2635 


1 1 


2 


2 


3 


4 


4 


5 


6 


.43 


2692 2698 


2704 


2710 


2716 


2723 


2729 


2735 


2742 


2748 


1 1 


2 


3 


3 


4 


4 


5 


6 


.44 


27541 2761 


2767 


2773 


2780 


2786 


2793 


2799 


2805 


2812 


1 1 


2 


3 


3 


4 


4 


5 


6 


.45 


2818 2825 


2831 


2838 


2844 


2851 


2858 


2864 


2871 


2877 


1 1 


2 


3 


3 


4 


5 


5 


6 


.46 


2884 2891 


2897 


2904 


2911 


2917 


2924 


2931 


2938 


2944 


1 1 


2 


3 


3 


4 


5 


5 


6 


47 


2951 2958 


2965 


2972 


2979 


2985 


2992 


2999 


3006 


3013 


1 1 


2 


3 


3 


4 


5 


5 


6 


.48 


3020 3027 


3034 


3041 


3048 


3055 


3062 


3069 


3076 


3083 


1 1 


2 


3 


4 


4 


5 


6 


6 


.49 1 30901 3097 


3105 


3112 


3119 


3126 


3133 


3141 


3148 


3155 


1 1 


2 


3 


4 


4 


5 


6 


_1 



ANTILOGARITHMS. 

1 


J. a 






















Proportional Parts. 


II 





1 


2 


3 


4 


5 


6 


7 


8 


9 




II 




1 














.50 


3162 




















1 


1' 

I 


2 


4 5 

3 


6 

4 4 


7 
5 


8 
6 


9 
7 


3170 


3177 


3184 


3192 


3199 


3206 


3214 


3221 


3228 


.51 


3236 


3243 


3251 


3258 


3266 


3273 


3281 


3289 


3296 


3304 




2 


2 


3 


4 5 


5 


6 


7 


.52 


3311 


3319 


3327 


3334 


3342 


3350 


3357 


3365 


3373 


3381 




2 


2 


3 


4 5 


5 


6 


7 


.53 


3388 


3396 


3404 


3412 


3420 


3428 


3436 


3443 


3451 


3459 




2 


2 


3 


4 5 


6 


6 


7 


.54 


3467 


3475 


3483 


3491 


3499 


3508 


3516 


3524 


3532 


3540 




2 


2 


3 


4 5 


6 


6 


7 


.55 


3548 


3556 


3565 


3573 


3581 


3589 


3597 


3606 


3614 


3622 




2 


2 


3 


4 5 


6 


7 


7 


.56 


3631 


3639 


3648 


3656 


3664 


3673 


3681 


3690 


3698 


3707 




2 


3 


3 


4 5 


6 


7 


8 


.57 


3715 


3724 


3733 


3741 


3750 


3758 


3767 


3776 


3784 


3793 




2 


3 


3 


4 5 


6 


7 


8 


.58 


3802 


3811 


3819 


3828 


3837 


3846 


3855 


3864 


3873 


3882 




2 


3 


4 


4 5 


6 


7 


8 


.59 
.60 


3890 
3981 


3899 
3990 


3908 


3917 


3926 
4018 


3936 
4027 


3945 
4036 


3954 
4046 


3963 
4055 


3972 
4064 




2 
2 


3 
3 


4 
4 


5 5 
5 6 


6 
6 


7 
7 


8 
8 


3999 


4009 


.61 


4074 


4083 


4093 


4102 


4111 


4121 


4130 


4140 


4150 


4159 




2 


3 


4 


5 6 


7 


8 


9 


.62 


4169 


4178 


4188 


4198 


4207 


4217 


4227 


4236 


4246 


4256 




2 


3 


4 


5 6 


7 


8 


9 


.63 


4266 


4276 


4285 


4295 


4305 


4315 


4325 


4335 


4345 


4355 




2 


3 


4 


5 6 


7 


8 


9 


.64 


4365 


4375 


4385 


4395 


4406 


4416 


4426 


4436 


4446 


4457 




2 


3 


4 


5 6 


7 


8 


9 


.65 


4467 


4477 


4487 


4498 


4508 


4519 


4529 


4539 


4550 


4560 




2 


3 


4 


5 6 


7 


8 


9 


.66 


4571 


4581 


4592 


4603 


4613 


4624 


4634 


4645 


4656 


4667 




2 


3 


4 


5 6 


7 


9 


10 


.67 


4677 


4688 


4699 


4710 


4721 


4732 


4742 


4753 


4764 


4775 




2 


3 


4 


5 7 


8 


9 


10 


.68 


4786 


4797 


4808 


4819 


4831 


4842 


4853 


4864 


4875 


4887 




2 


3 


4 


6 7 


8 


9 


10 


.69 


4898 


4909 


4920 


4932 


4943 


4955 


4966 


4977 


4989 


5000 




2 


3 


5 


6; 7 


8 


9 


10 


.70 


5012 


5023 


5035 


5047 


5058 


5070 


5082 


5093 


5105 


5117 




2 


4 


5 


6^ 7 


8 


9 


11 


.71 


5129 


5140 


5152 


5164 


5176 


5188 


5200 


5212 


5224 


5236 




2 


4 


5 


6 7 


8 


10 


U 


.72 


5248 


5260 


5272 


5284 


5297 


5309 


5321 


5333 


5346 


5358 




2 


4 


5 


6 7 


9 


10 


11 


.73 


5370 


5383 


5395 


5408 


5420 


5433 


5445 


5458 


5470 


5483 




3 


4 


5 


6 8 


9 


10 


11 


.74 


5495 


5508 


5521 


5534 


5546 


5559 


5572 


5585 


5598 


5610 




3 


4 


5 


6 8 


9 


10 


12 


.75 


5623 


5636 


5649 


5662 


5675 


5689 


5702 


5715 


5728 


5741 




3 


4 


5 


7j 8 


9 


10 


12 


.76 


5754 


5768 


5781 


5794 


5808 


5821 


5834 


5848 


586L 


5875 




3 


4 


5 


71 8 


9 


11 


12 


.77 


5888 


5902 


5916 


5929 


5943 


5957 


5970 


5984 


5998 


6012 




3 


4 


5 


7 8 


10 


11 


12 


.78 


6026 


6039 


6053 


6067 


6081 


6095 


6109 


6124 


6138 


6152 




3 


4 


6 


7 8 


10 


11 


13 


.79 


6166 


6180 


6194 


6209 


6223 


6237 


6252 


6266 


6281 


6295 




3 


4 


6 


7 9 


10 


11 


13 


.80 


6310 


6324 


6339 


6353 


6368 


6383 


6397 


6412 


6427 


6442 


1 


3 


4 


6 


7 9 


10 


12 


13 


.81 


6457 


6471 


6486 


6501 


6516 


6531 


6546 


6561 


6577 


6592 


2 


3 


5 


6 


8 9 


11 


12 


14 i 


•82 


6607 


662' 


6637 


6653 


6668 


6683 


6699 


6714 


6730 


6745 


2 


3 


5 


6 


8 9 


11 


12 


14 ; 


.83 


6761 


6776 


6792 


6808 


6823 


6839 


6855 


6871 


6887 


6902 


2 


3 


5 


6 


8 9 


11 


13 


14 


.84 


6918 


6934 


6950 


6966 


6982 


6998 


7015 


7031 


7047 


7063 


2 


3 


5 


6 


8 10 


11 


13 


15 


.85 


7079 


7096 


7112 


7129 


7145 


7161 


7178 


7194 


7211 


7228 


2 


3 


5 




8 10 


12 


13 


15 


.86 


7244 


7261 


7278 


7295 


7311 


7328 


7345 


7362 


7379 


7396 


2 


3 


5 




8 10 


12 


13 


15 


.87 


7413 


7430 


7447 


7464 


7482 


7499 


7516 


7534 


7551 


7568 


2 


3 


5 




9 10 


12 


14 


16 


.88 


7586 


7603 


7621 


7638 


7656 


7674 


7691 


7709 


7727 


7745 


2 


4 


5 




9 11 


12 


14 


16 


.89 


7762 


7780 


7798 


7816 


7834 


7852 


7870 


7889 


7907 


7925 


2 


4 


5 




9 11 


13 


14 


16 


.90 


7943 


7962 


7980 


7998 


8017 


8035 


8054 


8072 


8091 


8110 


2 


4 


6 


7 


9 11 


13 


15 


17 


.91 


8128 


8147 


8166 


8185 


8204 


8222 


8241 


8260 


8279 


8299 


2 


4 


6 


8 


9 11 


13 


15 


17 


.92 


8318 


8337 


8356 


8375 


8395 


8414 


8433 


8453 


8472 


8492 


2 


4 


6 


8 


12 


14 


15 


17 


.93 


8511 


8531 


8551 


8570 


8590 


8610 


8630 


8650 


8670 


8690 


2 


4 


6 


8 


12 


14 


16 


18 


.94 


8710 


8730 


8750 


8770 


8790 


8810 


8831 


8851 


8872 


8892 


2 


4 


6 


8 


12 


14 


16 


18 


,95 


8913 


8933 


8954 


8974 


8995 


9016 


9036 


9057 


9078 


9099 


2 


4 


6 


8 


12 


15 


17 


1 

19 


.96 


9120 


9141 


9162 


9183 


9204 


9226 


9247 


9268 


9290 


9311 


2 


4 


6 


8 


1 13 


15 


17 


19 


.97 


9333 


9354 


9376 


9397 


9419 


9441 


9462 


9484 


9506 


9528 


2 


•* 


7 


9 


1,13 


15 


17 


20 


.98 


9550 


9572 


9594 


9616 


9638 


9661 


9683 


9705 


9727 


9750 


2 


4 


7 


9 


1|13 


16 


18 


20 


.99 


9772 


9795 


9817 


9840 


9863 


9886 


9908 


9931 


9954 


9977 


2 


5 


7: 91 


1 14 161 18! 20 || 



LOGARITHMS OF NUMBERS. 


Il 

10 




0000 


1 


2 

0086 


3 


4 


5 


6 


7 


8 

033- 


9 


Proportional Parts. 


1 


2 


3 


4 


5 


G 


7 


8 


9 


0043 


0128 


0170 


0212 


0253 


0294 


0374 


4 


8 


12 


17 


21 


25 


29 


33 


37 


11 


0414 


0453 


0492 


0531 


0569 


0607 


0645 


0682 


071£ 


0755 


4 


8 


11 


15 


19 


23 


26 


30 


34 


12 


0792 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


107S 


1106 


3 


7 


10 


14 


17 


21 


24 


28 


31 


13 


1139 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


13<l£ 


1430 


3 


6 


10 


13 


16 


19 


23 


26 


29 


14 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


170: 


1732 


3 


6 


9 


12 


15 


18 


21 


24 


27 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


3 


6 


8 


11 


14 


17 


20 


22 


25 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


2233 


2279 


3 


5 


8 


11 


13 


16 


18 


21 


24 


17 


2304 


2330 


2355 


2380 


2405 


2430 


2455 


2480 


250-1 


2529 


2 


5 


7 


10 


12 


15 


17 


20 


22 


18 


2553 


2577 


2601 


2625 


264S 


2672 


2695 


2718 


2742 


2765 


2 


5 


7 


9 


12 


14 


16 


19 


21 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


2 


4 


7 


9 


11 


13 


16 


18 


20 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


2 


4 


6 


8 


11 


13 


15 


17 


19 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


2 


4 


6 


8 


10 


12 


14 


16 


18 


22 


3424 


3444 


3464 


3483 


3502 


3522 


3541 


3560 


3579 


3598 


2 


4 


6 


8 


10 


12 


14 


15 


17 


23 


3617 


3636 


3655 


3674 


3692 


3711 


3729 


3747 


3766 


3784 


2 


4 


6 


7 


9 


11 


13 


15 


17 


24 


3802 


3820 


3838 


3856 


3874 


3892 


3909 


3927 


3945 


3962 


2 


4 


5 


7 


9 


11 


12 


14 


16 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


2 


3 


6 


7 


9 


10 


12 


14 


15 


26 


4150 


4166 


4183 


4200 


4216 


4232 


4249 


4265 


42S1 


4298 


2 


3 


5 


7 


8 


10 


11 


13 


15 


27 


4314 


4330 


4346 


4362 


4378 


4393 


4409 


4425 


4440 


4456 


2 


3 


5 


6 


8 


9 


11 


13 


14 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564 


4579 


4594 


4609 


2 


3 


5 


6 


8 


9 


11 


12 


14 


29 


4624 


4639 


4654 


4669 


4683 


4698 


4713 


4728 


4742 


4757 




3 


4 


6 


7 


9 


10 


12 


13 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 




3 


4 


6 


7 


9 


10 


11 


13 


31 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


5011 


5024 


5038 




3 


4 


6 


7 


8 


10 


11 


12 


32 


5051 


5065 


5079 


5092 


5105 


5119 


5132 


5145 


5159 


5172 




3 


4 


5 


7 


8 


9 


11 


12 


33 


5185 


5198 


5211 


5224 


5237 


5250 


5263 


5276 


5289 


5302 




3 


4 


5 


6 


8 


9 


10 


12 


34 


5315 


5328 


5340 


5353 


5366 


5378 


5391 


5403 


5416 


5428 




3 


4 


5 


6 


8 


9 


10 


11 


35 


5441 


5453 


5465 


5478 


5490 


5502 


5514 


5527 


5539 


5551 




2 


4 


5 


6 


7 


9 


10 


11 


36 


5563 


5575 


5587 


5599 


5611 


5623 


5635 


5647 


5658 


5670 




2 


4 


5 


6 


7 


8 


10 


11 


37 


5682 


5694 


5705 


5717 


5729 


5740 


5752 


5763 


5775 


5786 




2 


3 


5 


6 


7 


8 


9 


10 


38 


5798 


5809 


5821 


5832 


5843 


5855 


5866 


5S77 


5888 


5899 




2 


3 


5 


6 


7 


8 


9 


10 


39 


5911 


5922 


5933 


5944 


5955 


5966 


5977 


5988 


5999 


6010 




2 


3 


4 


5 


7 


8 


9 


10 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 




2 


3 


4 


5 


6 


8 


9 


10 


41 


6128 


6138 


6149 


6160 


6170 


6180 


6191 


6201 


6212 


6222 




2 


3 


4 


5 


6 


7 


8 


9 


42 


6232 


6243 


6253 


6263 


6274 


6284 


6294 


6304 


6314 


6325 




2 


3 


4 


5 


6 


7 


8 


9 


43 


6335 


6345 


6355 


6365 


6375 


6385 


6395 


6405 


6415 


6425 




2 


3 


4 


5 


6 


7 


8 


9 


44 


6435 


6444 


6454 


6464 


6474 


6484 


6493 


6503 


6513 


6522 




2 


3 


4 


5 


6 


7 


8 


9 


45 


6532 


6542 


6551 


6561 


6571 


6580 


6590 


6599 


6609 


6618 




2 


3 


4 


5 


6 


7 


8 


9 


46 


6628 


6637 


6646 


6656 


6665 


6675 


6684 


6693 


6702 


6712 




2 


3 


4 


5 


6 


7 


7 


8 


47 


6721 


6730 


6739 


6749 


6758 


6767 


6776 


6785 


67S4 


6803 




2 


3 


4 


5 


5 


6 


7 


8 


48 


6812 


6821 


6830 


6839 


6848 


6857 


6866 


6875 


688-1 


6893 




2 


3 


4 


4 


5 


6 


7 


8 


49 


6902 


6911 


6920 


6928 


6937 


6946 


6955 


6964 


6972 


6981 




2 


3 


4 


4 


5 


6 


7 


8 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 




2 


3 


3 


4 


5 


6 


7 


8 


51 


7076 


7084 


7093 


7101 


7110 


7118 


7126 


7135 


7143 


7152 




2 


3 


3 


4 


5 


6 


7 


8 


52 


7160 


7168 


7177 


7185 


7193 


7202 


7210 


7218 


7226 


7235 




2 


2 


3 


4 


5 


6 


7 


7 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 


7300 


730S 


7316 




2 


2 


3 


4 


5 


6 


6 


7 


54 


7324 


7332 


7340 


7348 


7356 


7364 


7372 


7380 


7388 


7396 




2 


2 


3 


4 


5 


_1 


6 


7 



LOGARITHMS OF NUMBERS 












»>5 





1 


2 


3 


4 


5 


6 


7 


8 


9 


Proportional Parts. 


I 


2 


3 


4 


s 


6 


7 


8 


9 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 




2 


2 


3 


4 


5 


5 


6 


7 


56 


7482 


7490 


7497 


7505 


7513 


7520 


7528 


7536 


7543 


7551 




2 


2 


3 


4 


5 


5 


6 


7 


57 


7559 


7566 


7574 


7582 


7589 


7597 


7604 


7612 


7619 


7627 




2 


2 


3 


4 


5 


5 


6 


7 


58 


7634 


7642 


7649 


7657 


7664 


7672 


7679 


7686 


7694 


7701 






2 


3 


4 


4 


5 


6 


7 


59 


7709 


7716 


7723 


7731 


7738 


7745 


7752 


7760 


7767 


7774 






2 


3 


4 


4 


5 


6 


7 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 






2 


3 


4 


4 


5 


6 


6 


61 


7853 


7860 


7868 


7875 


7882 


7889 


7896 


7903 


7910 


7917 






2 


3 


4 


4 


5 


6 


6 


62 


7924 


7931 


7938 


7945 


7952 


7959 


7966 


7973 


7980 


7987 






2 


3 


3 


4 


5 


6 


6 


63 


7993 


8000 


8007 


8014 


8021 


8028 


8035 


8041 


8048 


8055 






2 


3 


3 


4 


5 


5 


6 


64 


8062 


8069 


8075 


8032 


8089 


8096 


8102 


8109 


8116 


8122 






2 


3 


3 


4 


5 


5 


6 


65 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 






2 


3 


3 


4 


5 


5 


6 


66 


8195 


8202 


8209 


8215 


8222 


8228 


8235 


8241 


8248 


8254 






2 


3 


3 


4 


5 


5 


6 


67 


8261 


8267 


8274 


8280 


8287 


8293 


8299 


8306 


8312 


8319 






2 


3 


3 


4 


5 


5 


6 


68 


8325 


8331 


8338 


8344 


8351 


8357 


8363 


8370 


8376 


8382 






2 


3 


3 


4 


4 


5 


6 


69 


8388 


8395 


8401 


8407 


8414 


8420 


8426 


8432 


8439 


8445 






2 


2 


3 


4 


4 


5 


6 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 






2 


2 


3 


4 


4 


5 


6 


71 


8513 


8519 


8525 


8531 


8537 


8543 


8549 


8555 


8561 


8567 






2 


2 


3 


4 


4 


5 


5 


72 


8573 


8579 


8585 


8591 


8597 


8603 


8609 


8615 


8621 


8627 






2 


2 


3 


4 


4 


5 


5 


73 


8633 


8639 


8645 


8651 


8657 


8663 


8669 


8675 


8681 


8686 






2 


2 


3 


4 


4 


5 


5 


74 


8692 


8698 


8704 


8710 


8716 


S722 


8727 


8733 


8739 


8745 






2 


2 


3 


4 


4 


5 


5 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 






2 


2 


3 


3 


4 


5 


5 


76 


8808 


8814 


8820 


8825 


8831 


8837 


8842 


8848 


8854 


8859 






2 


2 


3 


3 


4 


5 


5 


77 


8865 


8871 


8876 


8882 


8887 


8893 


8899 


8904 


8910 


8915 






2 


2 


3 


3 


4 


4 


5 


78 


8921 


8927 


8932 


8938 


8943 


8949 


8954 


8960 


8965 


8971 






2 


2 


3 


3 


4 


4 


5 


79 


8976 


8982 


8987 


8993 


8998 


9004 


9009 


9015 


9020 


9025 






2 


2 


3 


3 


4 


4 


5 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 






2 


2 


3 


3 


4 


4 


5 


81 


9085 


9090 


9096 


9101 


9106 


9112 


9117 


9122 


9128 


9133 






2 


2 


3 


3 


4 


4 


5 


82 


9138 


9143 


9149 


9154 


9159 


9165 


9170 


9175 


9180 


9186 






2 


2 


3 


3 


4 




5 


83 


9191 


9196 


9201 


9206 


9212 


9217 


9222 


9227 


9232 


9238 






2 


2 


3 


3 


4 




5 


84 


9243 


9248 


9253 


9258 


9263 


9269 


9274 


9279 


9284 


9289 






2 


2 


3 


3 


4 




5 


85 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 






2 


2 


3 


3 


4 




5 


86 


9345 


9350 


9355 


9360 


9365 


9370 


9375 


9380 


9385 


9390 






2 


2 


3 


3 


4 




5 


87 


9395 


9400 


9405 


9410 


9415 


9420 


9425 


9430 


9435 


9440 









2 


2 


3 


3 






88 


9445 


9450 


9455 


9460 


9465 


9469 


9474 


9479 


9484 


9489 









2 


2 


3 


3 






89 


9494 


9499 


9504 


9509 


9513 


9518 


9523 


9528 


9533 


9538 









2 


2 


3 


3 






90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 









2 


2 


3 


3 






91 


9590 


9595 


9600 


9605 


9609 


9614 


9619 


9624 


9628 


9633 









2 


2 


3 


3 






92 


9638 


9643 


9647 


9652 


9657 


9661 


9666 


9671 


9675 


9680 









2 


2 


3 


3 






93 


9685 


9689 


9694 


9699 


9703 


9708 


9713 


9717 


9722 


9727 









2 


2 


3 


3 






94 


9731 


9736 


9741 


9745 


9750 


9754 


9759 


9763 


9768 


9773 









2 


2 


3 


3 






95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 









2 


2 


3 


3 






96 


9823 


9827 


9832 


9836 


9841 


9845 


9850 


9854 


9859 


9863 









2 


2 


3 


3 






97 


9868 


9872 


9877 


9881 


9886 


9890 


9894 


9899 


9903 


9908 









2 


2 


3 


3 






98 


9912 


9917 


9921 


9926 


9930 


9934 


9939 


9943 


9948 


9952 









2 


2 


3 


3 






99 


9956 


9961 


9965 


9969 


9974 


9978 


9983 


9987 


9991 


9996 







_i 


2 


2 


3 


3 


3 







niH isdVHO IV ON do AiisyaAiNn